Plug $(0,n,0,\ldots,0)$ to see $\min g= 0$.
For the maximum, consider $n=2,3$ first, where the problem is immediately solved by AM-GM. The case $n=4$ is a bit more complicated, but not too hard. In the sequel we assume that $n\ge 5$. Since $g$ is continuous and is defined on a compact set, the maximum $g_0$ is attained at some point $(y_1,y_2,\ldots,y_n)$.
Claim. $y_3=\cdots=y_{n-2}$.
Proof. Suppose $y_k\not=y_{k+1}$ for some $3\le k\le n-3$. Then $g(\ldots,\frac{y_k+y_{k+1}}2,\frac{y_k+y_{k+1}}2,\ldots)>g_0$, a contradiction.
The following statement will be used frequently:
Claim. $n^2(n-5)/(n-4)^2\le(n+1)^2/4$, for $n\ge 5$.
Proof. For $n\ge 8$ it's plain since $n-5<n-4$ and $n-4\ge 4$. For $n=5,6,7$ we may check by direct computation.
Now let $y_3=u$, then
$$
g_0=(y_2+y_{n-1})u+y_1y_2+y_{n-1}y_n+y_1+y_n+(n-5)u^2
$$
If $y_2\not=y_{n-1}$ and $\min\{y_1,y_n\}\not=0$, then $(y_1,y_n)\mapsto(y_1+\delta,y_n-\delta)$ will increase $g$ for some $\delta$ (maybe negative). Hence $y_2\not=y_{n-1}\Rightarrow\min\{y_1,y_n\}=0$. Similarly $y_1\not=y_n\Rightarrow\min\{y_2,y_{n-1}\}=0$. Now there are several cases:
Case 1. $y_2=y_{n-1}, y_1=y_n$. If $y_1=y_2=0$, then
$$g_0=\frac{n^2(n-5)}{(n-4)^2}$$
which is no more than $(n+1)^2/4$ established in Case 2 and Case 3, a contradiction. Otherwise, if $y_1=y_n\not=0$, then by a perturbation $(y_1,y_n)\mapsto(y_1+\delta,y_n-\delta)$ the value of $g$ leaves unchanged, and the problem is reduced to Case 3; if $y_2=y_{n-1}\not=0$, it can be similarly reduced to Case 2.
Case 2. $y_2\not=y_{n-1}$. WLOG we may assume $y_n=0$.
Subcase 2.1 If $y_1\not=0$, we have $y_1>y_n$, hence $y_{n-1}=0$ (otherwise $(y_2,y_{n-1})\mapsto(y_2+\delta,y_{n-1}-\delta)$ will increase $g$ for $\delta>0$). So
\begin{align}
g_0&=y_2u+y_1y_2+y_1+(n-5)u^2\\
&=\frac{y_2(n-y_1-y_2)}{n-4}+y_1y_2+y_1+\frac{(n-5)(n-y_1-y_2)^2}{(n-4)^2}
\end{align}
The last expression is a convex quadratic function in $y_1$, thus attains its maximum at $y_1=0$ or $y_1=n-y_2$. When $y_1=0$,
\begin{align}
&\frac{y_2(n-y_2)}{n-4}+\frac{(n-5)(n-y_2)^2}{(n-4)^2}\\
=&-\frac{(n-y_2)^2}{(n-4)^2}+\frac{n(n-y_2)}{n-4}\\
\le&\cases{(5/2)^2,\phantom{n^2(n-5)/(n-4)^2}n=5\\n^2(n-5)/(n-4)^2,\phantom{(5/2)^2}n\ge 6}
\end{align}
and when $y_1=n-y_2$ we have
$$(n-y_2)y_2+n-y_2\le \frac{(n+1)^2}{4}$$
Now we conclude that(note that $(5/2)^2<(5+1)^2/4$)
$$g_0\le\max\left\{\frac{n^2(n-5)}{(n-4)^2},~\frac{(n+1)^2}4\right\}=\frac{(n+1)^2}{4}$$
Subcase 2.2 If $y_1=0$, then
\begin{align}
g_0&=(y_2+y_{n-1})u+(n-5)u^2\\
&=\Big(n-(n-4)u\Big)u+(n-5)u^2\\
&=-u^2+nu\le\frac{n^2}{4}<\frac{(n+1)^2}4
\end{align}
which is strictly smaller than Subcase 2.1, hence cannot be a maximum.
Case 3. $y_1\not=y_n$. WLOG we may assume $y_{n-1}=0$.
- Subcase 3.1 If $y_2\not=0$, by the same reasoning in Subcase 2.1 we have $y_n=0$. Now it's reduced to Subcase 2.1, where $y_{n-1}=y_n=0$.
- Subcase 3.2 If $y_2=0$, then
\begin{align}
g_0&=y_1+y_n+(n-5)u^2\\
&=n-(n-4)u+(n-5)u^2\\
&\le\max\left\{n,~\frac{n^2(n-5)}{(n-4)^2}\right\}\\
&\le\max\left\{\frac{(n+1)^2}{4},~\frac{n^2(n-5)}{(n-4)^2}\right\}\\
&=\frac{(n+1)^2}{4}
\end{align}
We may conclude that for $n\ge 5$, $g_0\le(n+1)^2/4$. And in fact $g_0=(n+1)^2/4$ since $g(\frac{n+1}2,\frac{n-1}2,0,0,\ldots,0)=(n+1)^2/4$.
Combining these results with the the case $n=3,4$, we may say that the maximum is $(n+1)^2/4$ for $n\ge 3$.
P.S. It's also worth mentioning that Lagrange multiplier method, mentioned in the comment, is not well applicable here. The result shows the extremal point is not a solution to the Lagrange multiplier equation.
