Let $G$ be a discrete group and let $M$ be a $G$module. Assume that I have a resolution $$\cdots \rightarrow M_1 \rightarrow M_0 \rightarrow M \rightarrow 0$$ of $M$ by $G$modules (with no further assumptions on the $M_k$; certainly they do not need to be free or acyclic or anything). What is the relationship between $H_{\ast}(G;M)$ and the abelian groups $H_{\ast}(G;M_k)$? It feels like there should be some kind of spectral sequence here, but I'm having trouble sorting out the relevant homological algebra.
This is an example of the hyperhomology spectral sequences in the case of chain complexes of $G$modules. See chapter 5.7 (especially proposition 5.7.6) of Weibel's book "An introduction to homological algebra". More precisely, let $P_{\ast\ast}$ be a Cartan–Eilenberg resolution of the chain complex of $G$modules $M_\ast$. Then each of the two spectral sequences (coming from the horizontal and vertical filtrations) associated with the double chain complex $P_{\ast\ast}/G$ will converge to the homology of the total complex of $P_{\ast\ast}/G$. The $E^2_{p,q}$term of the first spectral sequence is clearly $H_p(H_q(G;M_\ast))$, i.e. the $p$th homology of the chain complex $H_q(G;M_\ast)$. On the other hand, the $E^2_{p,q}$term of the second spectral sequence is $H_p(G;H_q(M_\ast))$. Since $H_q(M_\ast)$ is $M$ if $q=0$ and is trivial otherwise, we conclude that the second spectral sequence degenerates at the $E^2$ page, and its limit is canonically identified with $H_\ast(G;M)$. Therefore, the first spectral sequence converges to $H_\ast(G;M)$: $$ H_p(H_q(G;M_\ast)) \Longrightarrow H_{p+q}(G;M) $$ 

