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 $p=1$.

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:

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