There may be a geometric but partial answer to your question. This is an idea I learnt from Dennis Sullivan. As we know, passing from X$X$ to its universal cover X'$\widetilde{X}$ kills the fundamental group. Now, by Hurewicz we can assume that H_2(X')=\pi_2(X')$H_2(\widetilde{X})=\pi_2(\widetilde{X})$, whence killing H_2(X')$H_2(\widetilde{X})$ suffices. If we assume H_2(X')$H_2(\widetilde{X})$ is torsion free then each generator \alpha_i of H_2(X')$\alpha_i\in H_2(\widetilde{X})$ corresponds to a circle bundle E_i$E_i$ over X'$\widetilde{X}$, i.e., H_2(E)=H_2(X')/\Z\alpha_i$H_2(E_i)=H_2(\widetilde{X})/\mathbb{Z}\alpha_i$. Thus, if X'$\widetilde{X}$ was a manifold of dimension n$n$ and H_2(X')$H_2(\widetilde{X})$ was free of rank k$k$ then taking successive circle bundles we get a manifold E$E$ of dimension n+k$n+k$. This has the same higher homotopy groups (\pi_i$\pi_i$ for i>2$i>2$) as that of X$X$. The example given by Reid Barton is an illustration of this. However, for manifolds this is as far as you can go since killing even the free part of \pi_3(X')$\pi_3(\widetilde{X})$ (or, equivalently the free part of H_3(E)$H_3(E)$) requires bundles over E$E$ with fibre CP^\infty$\mathbb{CP}^\infty$, which lands us outside the realm of finite dimensional manifolds.
