Random Walks in $Z^2$/$Z^2$-intrinsic characterization of Euclidean distance Part II For some context see Random Walks in $Z^2$/$Z^2$-intrinsic characterization of Euclidean distance
As per Noah's answer and JBL's comment this was false as stated. However, I think the following reformulation is interesting. 
As before we consider a random walk on $\mathbb{Z}^2$ where a particle either stays at its vertex or moves to a neighbor with probability 1/5. We start the process with a particle at the origin.  For $x \in \mathbb{Z}^2$ we let $p_n(x)$ denote the probability that we find the particle at $x$ after $n$ iterations. Let $\left|\cdot\right|$ denote the Euclidean distance of two points in $\mathbb{Z}^2$ via the standard embedding of $\mathbb{Z}^2 \subset \mathbb{R}^2$. 
Now for the reformulated question: For each $n$, let $C_n$ be the supremum over all $C > 0$ so that for all $x,y \in \mathbb{Z}^2$ we have
$$
\text{$\left|x\right|, \left|y\right| \leq C$ and $\left|x\right| \leq \left|y\right| \Rightarrow p_n(x) \geq p_n(y)$. }
$$
Does $\lim_{n\to\infty} C_n = \infty$? If so, how fast does this diverge? 
EDIT: As per George Lowther's comment, I now find it quite probable that $\lim\inf_{n\to\infty} C_n \leq 5$ if not $C_n = 5$ for all large $n$.
A natural attempt to salvage the question is the following: For each $n$, let $\tilde{C}_n$ be the supremum over all $C > 0$ so that for all $x,y \in \mathbb{Z}^2$ we have 
$$
\text{$\left|x\right|, \left|y\right| \leq C$ and $\left|x\right| < \left|y\right| \Rightarrow p_n(x) > p_n(y)$. }
$$
Again we ask if $\lim_{n\to\infty} \tilde{C}_n = \infty$ and if so, how fast this diverges.
 A: By Donsker's theorem, this should converge to a Brownian motion in the scaling limit.  This means that the shapes Robby McKilliam plotted will converge to a circle (when properly scaled), since the distribution of Brownian motion is rotationally invariant.  Since the probability of moving from the current position is only 1/5 instead of 1, the time of the process will be slowed by a factor of 5, hence the radius of your limiting shapes will grow like $\sqrt{t/5}$ instead of $\sqrt{t}$.
A: I don't have the answer but I figured I would give you the results of a few quick experiments.
Here is what things look like when $n = 5$
 (source)
and when $n = 10$
 (source)
and when $n = 50$
 (source)
and when $n = 1000$
 (source)
The colour represents the probability, red being large, blue being small.  The actual colours are assigned according to the log of the probability. To generate these I used the following matlab
M = [ 0  1/5  0;
     1/5 1/5 1/5;
      0  1/5  0 ];
B = [1];
n = 50;
for i = 1:n
    B = conv2(B,M);
end
colormap(jet(256));
imagesc([-n, n], [-n, n], log(B));

Provided that the `shape' close to the origin becomes sufficiently circular, then the answer to your question is positive.
