$H_2(X)$ is all about $\pi_1(X)$ and $\pi_2(X)$. If $\pi_2(X)$ is trivial (as for knot complements) then it is a functor of $\pi_1(X)$. 

Let $H_n(G)$ be $H_n(BG)$, the homology of the classifying space ($K(G,1)$). If $X$ is path-connected than there is a surjection $H_2(X)\to H_2(\pi_1(X))$ whose kernel is a quotient of $\pi_2(X)$, the cokernel of a map from $H_3(\pi_1(X))$ to the largest quotient of $\pi_2(X)$ on which the canonical action of $\pi_1(X)$ becomes trivial.

This $H_2(G)$ isn't anything like the next piece of the derived series after $H_1(G)=G^{ab}$, though. For example, if $G$ is abelian then $H_2(G)$ is the second exterior power of $H_1(G)$ (EDIT: so it can be nontrivial even though it knows no more than $H_1(G)$ does), while if $H_1(G)$ is trivial $H_2(G)$ is often nontrivial (EDIT: so, even when it does carry some more information than $H_1(G)$, it is not necessarily derived-series information). 

EDIT: The previous paragraph comes from looking at the integral homology Serre spectral sequence of $X\to K(\pi_1(X),1)$, where the homotopy fiber is the universal cover $\tilde X$. Since $H_1\tilde X=0$, the groups $E^\infty_{p,1}$ are trivial and we get an exact sequence
$$
0\to E^\infty_{0,2}\to H_2(X)\to E^\infty_{2,0}\to 0,
$$
therefore
$$
E^2_{3,0} \to E^2_{0,2}\to H_2(X)\to E^2_{2,0}\to 0.
$$
Since $H_2(\tilde X)=\pi_2(\tilde X)=\pi_2(X)$, this looks like
$$
H_3(\pi_1(X)) \to H_0(\pi_1(X);\pi_2(X))\to H_2(X)\to H_2(\pi_1(X))\to 0.
$$
The place to look for the rest of the derived series would be homology with nontrivial coefficients, for example homology of covering spaces.