Update:
My memory was quite blurry about this when I originally answered.
See Gonzáles-Acuña, Gordon, Simon, ``Unsolvable problems about higher-dimensional knots and related groups,'' L’Enseignement Mathématique (2) 56 (2010), 143-171.
They prove that any finitely presented group is a subgroup of the fundamental group of the complement of a closed orientable surface in the $4$-sphere, which is much better than I reported.
Original answer:
You most likely would like a finitely presented group, but this might be of interest anyway:
Let $S$ be a recursively enumerable non-recursive subset of the natural numbers and consider the group
$\langle \ a,b,c,d \ | \ a^iba^{-i} = c^idc^{-i} \ \mathrm{for}\ i \in S \rangle$
This has unsolvable word problem. See page 110 of Chiswell's book "A course in formal languages, automata and groups" available on google books (I think it's also in Baumslag's "Topics in Combinatorial Group Theory" but all my books are in boxes at the moment.)
This should be the fundamental group of the complement of a noncompact surface in $\mathbb{R}^4$. You do this in the usual way by beginning with the trivial link on four components and then drawing the movie of the surface in $\mathbb{R}^4$, band summing at each stage to make the conjugates of $b$ and $d$ equal.
I think you end up with a knotted disjoint union of planes. I remember doing this at some point in graduate school when C. Gordon asked me if there were any compact surfaces in the $4$-sphere whose complements have groups with unsolvable word problem.
Best Answer
$H_2(X)$ 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.