As Ricardo points out in the comments, there's an error in my sketched calculation below. I also didn't notice the requirement that $X$ should be finite, so the natural $BK$ fails on two counts! However, it seems possible that a presentation complex for $K$ would do the job. Stallings shows that $\pi_2$ of any complex with $\pi_1=K$ is infinitely generated as a $K$-module.

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I think you want to start with a famous example of Stallings, from the paper 'A finitely presented group whose 3-dimensional integral homology is not finitely generated'. Stallings constructs a finitely presented group $K$ with the property that `there is no projective resolution of $\mathbf{Z}$ over $\mathbf{Z}[K]$ which is finitely generated in dimension 3' (Corollary 1).

In fact, as observed by Bieri, $K$ can be realizes as an explicit subgroup of the direct product of three free groups, $G=F_2\times F_2\times F_2$: $K$ is the kernel of a map $G\to\mathbf{Z}$ that sends every generator to $1$. So $K$ defines an `explicit' 3-complex, namely a covering space of the natural $K(G,1)$. As $K$ is 3-dimensional and finitely presented, it follows from Corollary 1 that $H_3(K,\mathbf{Z}[K])$ is infinitely generated.

Stallings's paper was the starting point for many beautiful constructions. Highlights include the word of Bieri and Bestvina--Brady .