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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 $X$ to its universal cover $\widetilde{X}$ kills the fundamental group. Now, by Hurewicz we can assume that $H_2(\widetilde{X})=\pi_2(\widetilde{X})$, whence killing $H_2(\widetilde{X})$ suffices. If we assume $H_2(\widetilde{X})$ is torsion free then each generator $\alpha_i\in H_2(\widetilde{X})$ corresponds to a circle bundle $E_i$ over $\widetilde{X}$, i.e., $H_2(E_i)=H_2(\widetilde{X})/\mathbb{Z}\alpha_i$. Thus, if $\widetilde{X}$ was a manifold of dimension $n$ and $H_2(\widetilde{X})$ was free of rank $k$ then taking successive circle bundles we get a manifold $E$ of dimension $n+k$. This has the same higher homotopy groups ($\pi_i$ for $i>2$) as that of $X$. 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 $\pi_3(\widetilde{X})$ (or, equivalently the free part of $H_3(E)$) requires bundles over $E$ with fibre $\mathbb{CP}^\infty$, which lands us outside the realm of finite dimensional manifolds.