Talking about naive attempts, I thought maybe a simple solution along these lines could be found, but I couldn't:
I denote by $p_k(n,2k)$ p_k(n,2r)$ the probability that a shortest path from the origin to $(n,2k)$ (n,2r)$ contains the segment $(\lfloor \frac{n}{2}\rfloor,k)$ to $(\lfloor \frac{n}{2}\rfloor +1,k)$. A simple observation is that $p_k(n,2k)=p_k(n,0)$ (consider reflecting the path on the line $y=k$ in the region $x > \lfloor \frac{n}{2}\rfloor$). The idea is to show that $p_k(n,2k)$ is close to $p_0(n,0)$ for small $k$. One can do this maybe by considering a new rectangular grid spanned by $(1,\frac{2k}{n})$ and $(-\frac{2k}{n},1)$ (with suitable edge weight distribution) and trying to find the new $p_0'(n,0)$ which should be a good approximation of $p_k(n,2k)$.
Now if one can find a slowly decreasing function $f$ so that $p_k(n,2k)\approx f(p_0(n,0))$ in the range, say $|k|\le \sqrt{n}$ then $$1=\sum_{k=-\infty}^{\infty}p_k(n,0)=\sum_{k=-\infty}^{\infty}p_k(n,2k)\approx \int_{-\sqrt{n}}^{\sqrt{n}}f(p)dp \geq c\sqrt{n}p_0(n,0)$$ for some constant $c$. If $\lim_{n\to \infty}p_0(n,0)>0$ then the above inequality is obviously false for large enough $n$.
ETA: I realize this approach works if we were able to prove $$\liminf_{n\to \infty} p_{0}(n,0)=\liminf_{n\to \infty} p_k(n,0)$$ for any fixed $k$.
