Let me replace the property "increasing" by the weaker "non-decreasing". This does not change the problem too much, but it makes the computation less messy.
It is well-known that the best linear function $a+bx$ satisfies the equations
$$
\left(\sum_i 1\right)a + \left(\sum_i x_i\right)b = \sum_i y_i \\
\left(\sum_i x_i\right)a + \left(\sum_i x_i^2\right)b = \sum_i x_iy_i.
$$
(Differentiate the function
$$
f(a,b) = \sum_i(a+bx_i-y_i)^2 = (a,b,1) \cdot\left(
\sum_i
\begin{pmatrix} 1 \\ x_i \\ -y_i \end{pmatrix}
(1, x_i, -y_i)
\right)
\cdot\begin{pmatrix} 1 \\ a \\ b \end{pmatrix}
$$
to see that.)
In our case $x_i=i$, and the equations are
$$
n\cdot a + \frac{n(n-1)}2\cdot b = \sum_{i=0}^{n-1} y_i \\
\frac{n(n-1)}2\cdot a + \frac{n(n-1)(2n-1)}6\cdot b = \sum_{i=0}^{n-1} iy_i .$$
By elimianting the slope $b$, the $y$-intercept is
$$
a = \frac6{n(n+1)} \sum_{i=0}^n \left(\frac{2n-1}3-i\right) y_i.
$$
In the last sum, the coefficient $\frac{2n-1}3-i$ is nonnegative for $i\le \frac{2n-1}3$ and it is negative for $i\ge \frac{2n}3$. In order to obtain a lower bound, omit the nonnegative terms, and
apply the upper bound $y_i\le y_{n-1}$ where the coefficient is positive:
$$
a \ge \frac6{n(n+1)} \left(\sum_{\frac{2n-1}3 <i\le n-1} \bigg(\frac{2n-1}3-i\bigg)\right) y_{n-1}. \tag{*}
$$
This is the lowest possible of $a$ if $n$ and $y_{n-1}$ are fixed.
For a precise evaluation we should consider three cases depending on the residue class of $n$ modulo $3$.
In $(*)$ the terms are all negative, they form an arithmetic progression, the first element is $\tfrac13$, $\tfrac23$ or $1$, the last element is $\frac{2n-1}3-(n-1)=-\frac{n-2}3$,
so the average is not less than $\frac12\left(-\frac{n-2}3-1\right)=-\frac{n+1}6$. The number of terms is at most $n-\frac{2n}3=\frac{n}3$. Hence,
$$
a \ge \frac6{n(n+1)} \left( \frac{n}3 \cdot \frac{-(n+1)}6 \right) y_{n-1}
= -\frac13 y_{n-1}.
$$
(Hope I did not make too serious miscalculations....)
