If you consider an exact sequence of groups, it's a fibration of topological spaces, so you get a homotopy long exact sequence. So that makes me think $\pi_n$ behaves like a derived functor from topological abelian groups to abelian groups. So there should be a spectral sequence whose first page is $\pi_m (G_n)$ and whose second page is $\pi_m$ of cohomology of $G_n$.

Because the cohomology of the first page is the second page, I think this implies that cohomology of $\pi_m (G_n)$ is just equal to $\pi_m$ of cohomology of $G_n$ (if this derived functor heuristic is correct, and I'm not very good at spectral sequences so it might be false regardless).

If you consider an exact sequence of groups, it's a fibration of topological spaces, so you get a homotopy long exact sequence. So that makes me think $\pi_n$ behaves like a derived functor from topological abelian groups to abelian groups. In particular this is the derived functor of $\pi_0$.

So there is the spectral sequence for applying a derived functor to a complex whose first page whose second page is $\pi_m$ of cohomology of $G_n$.

However, I don't think the first page is $\pi_m(G_n)$. One can see this from an exact sequence, like $\mathbb Z/2 \to S^1 \to S^1$. All the homotopy groups of the cohomology are trivial, because it is exact. But $\pi_m(G_n)$ has many nonzero terms and you can't make them all vanish just by applying a differential.

I don't know enough about spectral sequences to say exactly what the relationship between these two objects is, but I believe it is a question primarily of homological algebra.