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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 X' kills the fundamental group. Now, by Hurewicz we can assume that H_2(X')=\pi_2(X'), whence killing H_2(X') suffices. If we assume H_2(X') is torsion free then each generator \alpha_i of H_2(X') corresponds to a circle bundle E_i over X', i.e., H_2(E)=H_2(X')/\Z\alpha_i. Thus, if X' was a manifold of dimension n and H_2(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(X') (or, equivalently the free part of H_3(E)) requires bundles over E with fibre CP^\infty, which lands us outside the realm of finite dimensional manifolds.