Straighten the board out into a long straight track, so we are asking what is the probability of eventually landing on square $95$. This can only happen if we land on square $94$ and then roll a $1$ or we land on square $93$ and then roll a $2$, and so on. Define
$$p_n=\begin{cases}0,&n<0\\1,&n=0\\\frac16\sum_{k=1}^6p_{n-k},&n>0
\end{cases}$$
Then $p_n$ gives the probability of landing on square $n$ for $n>0$.
I wrote a python script to evaluate this and got $$p_{95}=0.2857142857142553$$
I've been playing with this for a couple of days, and after an embarrassingly long time, I've solved the general problem.
A point moves on the real line according to the rolls of a fair $s$-sided die. It starts at the origin at time $t=0$. Thereafter, if it is at $x=k$ at time $t$, it moves to $x=k+r$ with probability $1/s$ for $r=1,2,\dots,s$. Let $p_n$ be the probability that the point lands at $x=n$ at some point in time.
I claim that $$\lim_{n\to\infty}p_n=\frac2{s+1}.\tag1$$ It's clear that $p_n$ is the solution to the linear difference equation $$p_n=\frac1s\sum_{k=1}^sp_{n-k}\tag2$$ with initial conditions $$p_n=0 \text{ for } n<0,\text{ and }p_0=1.$$
The characteristic function of $(2)$ is $$f(x)=x^s-\frac{x^{s-1}}s-\cdots-\frac xs-\frac1s$$ Let the roots of $f$ be $\theta_1=1,\theta_2,\dots,\theta_s$. I will show that $|\theta_k|<1$ for $1<k\leq s$, and that the solution to $(2)$ is $$p_n=\frac2{s+1}+\frac1{s+1}\sum_{k=2}^s\theta_k^n,\ n\geq0.\tag3$$ Of course, $(1)$ follows immediately.
Define $\pi_n=\sum_{k=1}^s\theta^k,\ n\geq0$, so that $(4)$ becomes $$p_n=\frac1{s+1}+\frac{\pi_n}{s+1},\ \ n\geq0\tag4$$
For $k=1,\dots,s$ let $e_k$ be the $k$th elementary symmetric function of the $\theta_k$. Vieta's formulas give
$$e_k=\frac{(-1)^{k+1}}s$$ and Newton's identities give $$
\begin{align}
\pi_n&=(-1)^{n-1}ne_n+\sum_{k=1}^{n-1}e_{n-k}\pi_k\\
&=\frac ns+\frac1s\sum_{k=1}^{n-1}\pi_k,\ 1\leq n\leq s\\\tag5
\end{align}$$
and
$$\begin{align}
\pi_n&=\sum_{k=n-s}^{n-1}(-1)^{n-1+k}e_{n-k}\pi_k\\
&=\frac1s\sum_{k=1}^s\pi_{n-k},\ n\geq s\\\tag6
\end{align}$$
Now it's easy to show that the $p_n$ defined in $(4)$ satisfy the difference equation. When $1\leq n\leq s$, $(5)$ gives
$$\begin{align}
p_n&=\frac1{s+1}+\frac1{s+1}\pi_n\\
&=\frac1{s+1}+\frac1{s+1}\left(\frac ns+\frac1s\sum_{k=1}^{n-1}\pi_k\right)\\
&=\frac1{s+1}+\frac n{s+1}+\frac1{s(s+1)}\sum_{k=1}^{n-1}((s+1)p_k-1)\\
&=\frac1{s+1}+\frac n{s+1}-\frac{n-1}{s+1}+\frac1s\sum_{k=1}^{n-1}p_k\\
&=\frac1s+\frac1s\sum_{k=1}^{n-1}p_k
=\frac1s\sum_{k=0}^{n-1}p_k=\frac1s\sum_{k=1}^sp_{n-k},
\end{align}$$
When $n>s$, $(6)$ gives
$$\begin{align}
p_n&=\frac1{s+1}(1+\pi_n)\\
&=\frac1{s+1}+\frac1{s+1}\left(\frac1s\sum_{k=1}^s\pi_{n-k}\right)\\
&=\frac1{s+1}+\frac1{s(s+1)}\left(\sum_{k=1}^s((s+1)p_{n-k}-1)\right)\\
&=\frac1{s+1}-\frac s{s(s+1)}+\frac1s\sum_{k=1}^sp_{n-k}=\frac1s\sum_{k=1}^sp_{n-k}
\end{align}$$
It remains to show $|\theta_k|<$ when $k\neq1$. We have $$
\begin{align}
sf(x) &=sx^s-x^{s-1}-\cdots-x-1\\&=(x-1)\sum_{k=0}^{s-1}(k+1)x^k\\&:=(x-1)g(x)\end{align}$$ and
$$\begin{align}
g(x)&=\frac d{dx}(x^s+x^{s-1}+\cdots+x+1)\\&=\frac d{dx}\frac{x^s-1}{x-1}\\&:=\frac d{dx}G(x)
\end{align}$$
The roots of $G$ are $s-1$ of the $s$th roots of unity, and by the Gauss-Lucas theorem the roots of $g$ lie in their convex hull. Note that a root of $g$ can only have modulus $1$ if it coincides with a zero of $G$, and we know that $G$ has only simple roots. $\square$
