As I mentioned in my comment - what you are suggesting implies that the probability of being at (3,4) is the same as being at (5,0) for all large n. That seems unlikely, and would guess that $C_n=5$ for n large.


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The answer to your modified question is yes, $\tilde C_n$ tends to infinity as n goes to infinity. Also, the rate is at least $\sqrt{\frac45n\log n}$. That is,
$$
\liminf_{n\to\infty}\frac{\tilde C_n}{\sqrt{\frac45n\log n}}\ge1.
$$
This can be proven by evaluating $p_n(x)$ to leading order in 1/n. (assuming I haven't made any errors below).

The idea is to note that you are repeatedly applying a linear operator,
$$
p_{n+1}=Lp_n,\ Lp(x) \equiv (p(x)+p(x-e_1)+p(x+e_1)+p(x-e_2)+p(x+e_2))/5
$$
where $e_1=(1,0)$, $e_2=(0,1)$. In finite dimensional spaces, you would solve this by decomposing $p_0$ into a sum of eigenvectors and for large n, the dominant term of $L^np_0$ will be that corresponding to the largest eigenvalue. In this case, we can diagonalize by applying a Fourier transform
$$
p_0(x)=1_{\lbrace x=0\rbrace}=\int_{-[\frac12,\frac12]^2}e^{2\pi ix\cdot u}\\,du.
$$
Noting that $e^{2\pi ix\cdot u}$ (as a function of x) is an eigenvector of L,
$$
Le^{2\pi ix\cdot u}=\left(\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)\right)e^{2\pi ix\cdot u}
$$
gives the following for $p_n$,
$$
p_n(x)=L^np_0(x)=\int_{[-\frac12,\frac12]^2}\left(\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)\right)^ne^{2\pi ix\cdot u}\\,du.
$$
The term inside the parentheses is less than 1 in absolute value everywhere away from the origin, so looks like a Dirac delta when  raised to a high power n.

The integral can be computed to leading order. After n steps, the standard deviation of the particle's distance from the origin grows of the order of $\sqrt{n}$, so most of its distribution is spread over an area of size the order of n. This means we expect $p_n(x)$ to go to zero at rate 1/n. Let's discard all terms which vanish faster than this.

The range of integration can be replaced by $[-\epsilon,\epsilon]$ for any $0<\epsilon<1/2$, creating an error only of order $e^{-cn}$ (some positive c). On such a range,
$$
\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)=\exp\left(-\frac{4\pi^2}{5}u^2+O(u^4)\right)
$$
Substituting into the integral and changing variables,
$$
\begin{align}
p_n(x)&=\int_{[-\epsilon,\epsilon]^2}\exp\left(-\frac{4n\pi^2}{5}\vert u\vert^2+O(n\vert u\vert^4)+2\pi ix\cdot u\right)\\,du+O(e^{-cn})\\\\
&=\frac{5}{8n\pi^2}\int_{[-\epsilon'\sqrt{n},\epsilon'\sqrt{n}]^2}\exp\left(-\frac12\vert v\vert^2+O(\vert v\vert^4/n)+i\sqrt{\frac{5}{2n}}x\cdot v\right)\\,dv+O(e^{-cn})
\end{align}
$$
Where $\epsilon'=\epsilon\sqrt{8\pi^2/5}$. Rearranging the v^4/n term in the exponential, it is possible to show that it only contributes of the order of 1/n to the integral. Integrals like $\int_{\epsilon\sqrt{n}}^\infty\exp(-x^2/2)\\,dx$ vanish at rate $\exp(-\epsilon^2n/2)/n$, giving
$$
\begin{align}
p_n(x)&=\frac{5}{8n\pi^2}\int_{\mathbb{R}^2}\exp\left(-\frac12\vert v\vert^2+i\sqrt{\frac{5}{2n}}x\cdot v\right)\\,dv+O(1/n^2)\\\\
&=\frac{5}{4n\pi}\exp\left(-\frac{5\vert x\vert^2}{4n}\right)+O(1/n^2)
\end{align}
$$
So, if $\vert x\vert\lt\vert y\vert$ then $p_n(x)\gt p_n(y)$ for large n. You can check from this expression that $\vert x\vert\lt c\sqrt{\frac45n\log n}$ (any $c\lt1$) is enough for the first term to dominate in the inequality.