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 $n = 5$

and when $n = 10$

and when $n = 50$

and when $n = 1000$

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.

3 deleted 131 characters in body

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 $n = 5$

and when $n = 10$

and when $n = 50$

and when $n = 1000$

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));


The trend is reasonably clear. It appears

Provided that the general shape' stays the same and we simply sample it more finely as $n$ increases. As the shape looks close to be circular around the origin becomes sufficiently circular, this lends evidence for an affirmative then the answer to your question is positive.

2 added 18 characters in body

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 $n = 5$

and when $n = 10$

and when $n = 50$

and when $n = 1000$

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(log(B))imagesc([-n, n], [-n, n], log(B));


The trend is reasonably clear. It appears that the general `shape' stays the same and we simply sample it more finely as $n$ increases. As the shape looks to be circular around the origin, this lends evidence for an affirmative answer to your question.

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