For an $n \times n$ grid, the probability of finding a path of length $n\sqrt{2}$ is $1/2^n = 2^{-n}$.

For a grid of size $(n,0)$ or $(0,n)$, the expected path length is $n$ with probability $p=1$. Let's call the expected path length $L(x,y)$

$$L(n,0)=L(0,n)=1 \cdot n = n$$



For a grid of size $(n,1)$ or $(1,n)$, the expected path length is $n+1$ if all possible diagonals face the incorrect way, and $n+\sqrt{2}$ if there exists at least one-diagonal facing the correct way to create a short-cut:

$$L(n,1)=L(1,n)=\binom{n}{1} \cdot \frac{1}{2^n} \cdot n + (1 - \binom{n}{1} \cdot \frac{1}{2^n}) \cdot (n+\sqrt{2} )$$

$$L(n,1)=L(1,n)= n + (1 - \binom{n}{1} \cdot \frac{1}{2^n}) \cdot \sqrt{2}$$

$$L(n,1)=L(1,n)= n + (\frac{2^n -1}{2^n}) \cdot \sqrt{2}$$

For a grid size $(n,2)$ or $(2,n)$, the expected shortest path length is $n+2$ if in all of the locations, there are no correct facing diagonals; $n+1+\sqrt{2}$ if the short-cut diagonal only occurs in the last square (top-most); or length $n+2\sqrt{2}$, if there is a short-cut diagonal in one of the first column's $n-1$ lower squares, and a short-cut diagonal in one of the second column's upper squares after the lower square.  

This could probably be written as a recursive formula to see what the limit yields, as $L(0,0)=0$, $L(1,0)=L(0,1)=1$, $L(1,1)=\sqrt(2)\cdot\frac{1}{2}+2\cdot\frac{1}{2}=1+\frac{\sqrt{2}}{2}$,