There is no uniform distribution over $[0,\infty),$ but you can assume
a uniform distribution over any finite number of values that you want.
So let's consider uniform distributions of each of the variables
$N_\mathrm{neg},$ $N_\mathrm{neut},$ and $N_\mathrm{pos}$
over all integers in the interval from a lower bound $L$ up to and including
an upper bound $U$, that is, the set of integers
$I = \{L, L+1, L+2, \ldots, U\},$ where $0 \leq L < U.$
The joint distribution of the three variables is uniform on
$I \times I \times I,$ a “cube” of points with integer
coordinates in three-dimensional space.
This joint distribution defines the distribution of a variable $score$
over the interval $[-1,1],$ where
$$
score = \frac{N_\mathrm{pos} - N_\mathrm{neg}}
{N_\mathrm{neg} + N_\mathrm{neut} + N_\mathrm{pos}}.
$$
To simplify the notation for the following calculations, define variables
$x,$ $y,$ and $z$ such that
\begin{align}
N_\mathrm{pos} &= (U - L)x + L,\\
N_\mathrm{neg} &= (U - L)y + L,\\
N_\mathrm{neut} &= (U - L)z + L
\end{align}
and also set $a = \frac{3L}{U - L}.$ Then
$$
score = \frac{x - y}{x + y + z + a}.
$$
By taking $(x,y,z)$ as coordinates of a point in three-dimensional space
instead of $(N_\mathrm{neg},N_\mathrm{neut},N_\mathrm{pos}),$
we transform the set of voting configurations to points on or inside
the unit cube $[0,1]\times[0,1]\times[0,1].$
Now consider the value of $score$ at each of the eight vertices of the cube:
\begin{array}{ccrcl}
(x,y,z) & \quad &&&\llap{score} \\
\hline
(0,0,0) && \dfrac{0-0}{0+0+0+a} &=& 0, \\
(1,0,0) && \dfrac{1-0}{1+0+0+a} &=& 1-\dfrac{a}{a+1}, \\
(0,1,0) && \dfrac{0-1}{0+1+0+a} &=& -\left(1-\dfrac{a}{a+1}\right), \\
(1,1,0) && \dfrac{1-1}{1+1+0+a} &=& 0, \\
(0,0,1) && \dfrac{0-0}{0+0+1+a} &=& 0, \\
(1,0,1) && \dfrac{1-0}{1+0+1+a} &=& \dfrac12\left(1-\dfrac{a}{a+2}\right),\\
(0,1,1) && \dfrac{0-1}{0+1+1+a} &=& -\dfrac12\left(1-\dfrac{a}{a+2}\right),\\
(1,1,1) && \dfrac{1-1}{1+1+1+a} &=& 0.
\end{array}
From these eight vertices, we can already get an intuition that the
values of $score$ will tend to cluster around zero;
at half of the vertices, $score = 0,$
and at two others $-\frac12 \leq score \leq \frac12.$
There are only two vertices at which $score$ can even get close to either
$-1$ or $1,$ and it cannot actually reach those values unless $a = 0$
(that is, unless $L = 0$).
For example, with $L=10$ and $U=25,$ the values of $score$ at the eight
vertices of the cube are
$0,$ $\frac13,$ $0,$ $\frac14,$ $-\frac13,$ $0,$ $-\frac14,$ and $0.$
In this example, all the values of $score$ will be in the range
$\left[-\frac13, \frac13\right].$
Now suppose $T$ is a positive number.
To estimate the proportion of the ordered triples
$(N_\mathrm{neg},N_\mathrm{neut},N_\mathrm{pos})$ for which the
score is greater than $T,$
we will first consider the case in which
$T < \frac12\left(1 - \frac{a}{a+2}\right).$
In this case, if we allow interpolation between points with
integer coordinates, we can find all points at which $score \geq T$
in the plane $z = v$ for any particular value of $v$
such that $0 \leq v \leq 1.$
These points lie in a triangle bounded by lines with equations
$(1-T)x - (1+T)y = T(v+a),$ $y = 0,$ and $x = 1.$
The vertices of the triangle are $(1,0,v),$
$\left(\frac{T(v+a)}{1-T},0,v\right),$ and
$\left(1,\frac{1 - T - T(v+a)}{1+T},v\right).$
It is a right triangle with legs $1 - \frac{T(v+a)}{1-T}$
and $\frac{1 - T - T(v+a)}{1+T}$;
its area is $\frac{(1 - T - T(v+a))^2}{2(1-T^2)}.$
Therefore the portion of the cube within which $score \geq T$ has volume
$$
\int_0^1 \frac{(1 - T - T(v+a))^2}{2(1-T^2)}\,dv
= \frac{(3a^2 + 9a + 7)T^2 - (6a + 9)T + 3}{6(1 - T^2)}.
$$
We want this to be one-third the volume of the cube,
that is, we want to set
$$ \frac{(3a^2 + 9a + 7)T^2 - (6a + 9)T + 3}{6(1 - T^2)} = \frac13. $$
Simplifying,
\begin{align}
(3a^2 + 9a + 7)T^2 - (6a + 9)T + 3 &= 2 - 2T^2,\\
(3a^2 + 9a + 9)T^2 - (6a + 9)T + 1 &= 0,\\
\end{align}
and solving this as a quadratic equation in $T,$
$$
T = \frac{6a + 9 \pm \sqrt{(6a + 9)^2 - 4(3a^2 + 9a + 9)(1)}}
{2(3a^2 + 9a + 9)}
= \frac{6 a + 9 \pm \sqrt{24 a^2 + 72 a + 45}}{6 a^2 + 18 a + 18}.
$$
The condition $T < \frac12\left(1 - \frac{a}{a+2}\right)$
dictates that the only possible solution is
$$T = \frac{6 a + 9 - \sqrt{24 a^2 + 72 a + 45}}{6 a^2 + 18 a + 18}.$$
Moreover, this solution works for all $a \geq 0,$ and since there can only
be one value of $T$ such that $\frac13$ of the volume of the cube consists of
points for which $score \geq T,$
we can rule out any solutions for the case
$T \geq \frac12\left(1 - \frac{a}{a+2}\right).$
Of course, the ratio of the number of possible discrete values of $(x,y,z)$
within this particular part of the cube to the total number of
possible discrete values of $(x,y,z)$
(corresponding to possible voting configurations)
is only approximately equal to the volume of that part.
The approximation is better when the number of configurations is very large.
For example, with $L=10,$ $U=25,$ we have $a = 2$ and we estimate
$T = \frac{21 - \sqrt{137}}{76} \approx 0.052796.$
According to the python script I wrote to count the voting configurations,
this gives us $1304$ neutral configurations (a little less than
$32\%$ of the total), and $1396$ each of positive and negative configurations
(a little more than $34\%$ of the total).
Also note that the value of $T$ depends on the particular values of
$L$ and $U$ that are chosen. Intervals in which $U$ is only a few percent
larger than $L$ will require small values of $T$;
intervals in which $L$ is a small fraction of $U$
(or in which $L=0$) will require larger values of $T.$
But the largest possible value of $T$ is something near
$\frac{9 - \sqrt{45}}{18} \approx 0.127322.$
Improved Estimate
A slightly better estimate of the desired value of $T$ may be obtained
by mapping the coordinates so that the vertices of the unit cube
correspond to fictional configurations with
$L - \frac12$ or $U + \frac12$ of each type of vote.
The actual configurations then would all map to points inside the cube,
never on the cube's surface.
We modify the formulas for $x,$ $y,$ and $z$ as follows:
\begin{align}
N_\mathrm{pos} &= (U - L + 1)x + L - \frac12,\\
N_\mathrm{neg} &= (U - L + 1)y + L - \frac12,\\
N_\mathrm{neut} &= (U - L + 1)z + L - \frac12.
\end{align}
We set $a = \frac{6L - 3}{2(U - L + 1)}.$ Then once again
$$
score = \frac{x - y}{x + y + z + a}.
$$
The calculation of the portion of the cube (by volume) in which
$score \geq T$ therefore proceeds as before;
the volume is $\frac13$ of the entire cube when
$$
T = \frac{6 a + 9 - \sqrt{24 a^2 + 72 a + 45}}{6 a^2 + 18 a + 18}.
$$
We use the modified value of $a,$ however.
For the example $L=10$ and $U=25$ we have
$a = \frac{6(10) - 3}{2(25 - 10 + 1)} = \frac{57}{32} = 1.78125,$
so
$$
T = \frac{3360 - 32 \sqrt{7094}}{11793} \approx 0.056370.
$$
Using this value of $T,$ my python script finds $1366$ neutral
configurations and $1365$ each of positive and negative configurations.
Supporting Code
Here's a python script that returns a list of numbers counting (respectively)
the NEG, NEUT, and POS decisions for all configurations of votes in which
the numbers of each of the three vote types range from
min_votes to max_votes inclusive, and T is the value of $T.$
def countdecisions(min_votes, max_votes, T):
count1 = 0
count2 = 0
count3 = 0
voting_range = range(min_votes, max_votes + 1)
for pos in voting_range:
for neg in voting_range:
for neut in voting_range:
score = getscore(pos, neg, neut)
if (score <= -T):
count1 += 1
elif (score >= T):
count3 += 1
else:
count2 += 1
return [count1, count2, count3]
Old Answer
The following was an attempt to answer the question as previously written.
Mathematically, when you write a limit like this,
$$
\lim_{\substack{N_\mathrm{neg} \to +\infty \\[0.2ex]
N_\mathrm{neut} \to +\infty \\[0.2ex] N_\mathrm{pos} \to +\infty}}
f(N_\mathrm{neg}, N_\mathrm{neut}, N_\mathrm{pos}),
$$
you are asking about what happens to
$f(N_\mathrm{neg}, N_\mathrm{neut}, N_\mathrm{pos})$
when all three of its parameters grow above all bounds,
and you are requesting an answer that will be valid in any situation
in which all three parameters grow above all bounds.
For example, a limit written in this way should have the same value if
the three parameters consistently satisfy
$N_\mathrm{neg} > 3(N_\mathrm{neut}) > 3(N_\mathrm{pos})$ as the numbers grow,
or if $N_\mathrm{neut} > 3(N_\mathrm{neg}) > 3(N_\mathrm{pos})$ instead,
just as long as none of the parameters $N_\mathrm{pos},$ $N_\mathrm{neg},$
or $N_\mathrm{neut}$ is bounded by any constant number.
It should be evident merely by looking at how you set up the formula
to see that there is no value of $T$ less than $1$ for which the limit
is even defined, let alone where it is equal to $1.$
By choosing different relative rates at which the parameters grow,
you can make the scores cluster as close to $-1$ as you want,
or as close to $0$ as you want,
so you can make either of the quantities $\mathrm{COUNT}(score \in [-1, -T])$
or $\mathrm{COUNT}(score \in [-T,T])$
grow much faster than the other.
Now, I would think that it would be a desirable feature that in almost all
cases where the users' votes are mostly positive, the score would end up
in the interval $[T,1],$ whereas in almost all cases where the positive
and negative votes were roughly equal, the score would end up in
the interval $[-T,T].$
If your computational methods consist of randomly generating votes with
equal likelihood of each of the three possible choices for each vote,
the law of large numbers says the score will tend to $0$ as the number of votes increases; that is, for any positive constant $T$ you choose,
the score will be in $[-T,T]$ after $N$ votes with a probability that
approaches $1$ as $N$ goes to infinity.
It seems to me that this is how a voting system ought to work in such a case.
If you really want the probability of scores in $[-1,-T],$ $[-T,T],$
and $[T,1]$ to be approximately equal after $N$ votes have been cast, however,
you can use a normal approximation of the distribution of scores.
Let $X_i$ be a random variable defined by
$$
X_i = \begin{cases} \phantom{-}1 & \text{if the $i$th vote is $POS$}, \\
\phantom{-}0 & \text{if the $i$th vote is $NEUT$}, \\
-1 & \text{if the $i$th vote is $NEG$}. \\ \end{cases}
$$
Then after $N$ votes have been cast,
$$
N_\mathrm{pos} - N_\mathrm{neg}
= X_1 + X_2 + X_3 + \cdots + X_N = \sum_{i=0}^N X_i.
$$
Assuming each kind of vote is equally likely to be cast next,
the variance of $X_i$ is $\frac23,$
and the variance of $\sum_{i=0}^N X_i$ is $\frac23N.$
The standard deviation is the square root of the variance,
$\sqrt{\frac23 N}.$
The mean of $\sum_{i=0}^N X_i$ is zero.
For large $N,$ the distribution of $\sum_{i=0}^N X_i$ approximates
a normal distribution.
A random variable with a normal distribution with mean zero
and standard deviation $\sigma$
is greater than approximately $0.430727\sigma$ with probability $\frac13.$
For large $N,$ then, we will find that
$$
\sum_{i=0}^N X_i > 0.430727\sqrt{\frac23 N} \approx 0.351687\sqrt{N}
$$
approximately $\frac13$ of the time.
The score you assign is equal to $\frac1N \sum_{i=0}^N X_i,$
so if you let $T = \frac{0.351687}{\sqrt{N}}$
(a function of the number of votes cast, not a constant),
you should find that the likelihoods of the score being in
$[-1,-T],$ $[-T,T],$ and $[T,1]$ are approximately equal.
