All we can say is that all paths are equally likely, so $\mathbb{P}[X(n)=x]=\frac{\text{# walks of length }n\text{ that terminate at }x}{4^n}$. As far as calculating the numerator, if $x=(a,b)$ and $s\geq |a|$, $n-s\geq |b|$, then the number of walks of length $n$ that terminate at $x$ and have $s$ horizontal movements are $\left(\begin{array}{c} s \\ \frac{s+|a|}{2}\end{array}\right)\left(\begin{array}{c} n-s \\ \frac{n-s+|b|}{2}\end{array}\right)$ (the first term is the number of ways you can space your movements in the $(a,0)$ direction with $s$ opportunities, the second is similar for vertical). So $\text{# walks of length }n\text{ that terminate at }x=\sum_{s=|a|}^{n-|b|}\left(\begin{array}{c} s \\ \frac{s+|a|}{2}\end{array}\right)\left(\begin{array}{c} n-s \\ \frac{n-s+|b|}{2}\end{array}\right)$, where we understand that the terms with fractional arguments are equal to 0. It's hard to get further than this without going round in circles, but this is usually enough to motivate a computer algorithm.