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.
The answer to your modified question is yes, $\tilde C_n$ tends to infinity as n goes to infinity. Edit: I'm not so sure now, there was a mistake in my original calculation. What you can say is that there is a positive constant $a$ such that the result holds as long as you restrict to $\vert x\vert^2\lt\vert y\vert^2-a$. Whether or not we can take $a=1$ (which would imply the full result, as $\vert x\vert^2-\vert y\vert^2\in\mathbb{Z}$) requires a more careful consideration of the second order terms in my calculation. In any case, the rate will be of the order of $\sqrt{n}$. That is, $\tilde C_n\ge b\sqrt{n}$ for some constant $b$.
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 that if $\vert x\vert\lt\vert y\vert-a$ for some constant a, $$ \begin{align} p_n(x)-p_n(y)&=\frac{5}{4n\pi}e^{-\frac{5\vert x\vert^2}{4n}}(1-e^{\frac{5(\vert x\vert^2-\vert y\vert^2)}{4n}})+O(1/n^2)\\\\ &\ge\frac{5}{4n\pi}e^{-\frac{5\vert x\vert^2}{4n}}(1-e^{-\frac{5a}{4n}})+O(1/n^2)\\\\ &\ge\frac{25}{4n^2\pi}\left(ae^{-\frac{5\vert x\vert^2}{4n}}+O(1)\right) \end{align} $$ To be certain that this is positive, we would need to evaluate the constant term in the final expression and compare to $a$.