gt.geometric topology - Homotopy groups of smooth manifolds?

For $d=3$ the homotopy groups can be pretty elaborate. Consider the connect-sum of some lens spaces. The universal cover embeds in $S^3$ as the complement of a cantor set (except for a few degenerate cases where you have $mathbb RP^3$ summands). So the homotopy-groups are pretty complicated ($pi_2$ is finitely generated over $pi_1$). You could probably make an argument that this is about the worst thing that can happen for the homotopy-groups of 3-manifolds.

You might want to phrase your question as a question about the Postnikov towers of manifolds. Eilenberg-Maclane spaces are rarely compact boundaryless manifolds.

edit: I guess another spin on your question could go like this. We know the fundamental groups of compact manifolds are all possible finitely presentable groups provided $n geq 4$. So is there a sense in which the homotopy-algebras of manifolds can be anything finitely presentable? Say, for example, $pi_2$. As a module over the group-ring of $pi_1$, are there any restrictions beyond being finitely generated? I suppose you could construct a compact $6$-manifold with $pi_2$ (almost) any finitely-presented thing over any finitely-presented $pi_1$ pretty much the exact same way $4$-manifolds with any finitely presented $pi_1$ are constructed. I think if $H_2(pi_1)$ is non-trivial you might run into problems following the analogous construction, in that $pi_2$ might strictly contain the $pi_2$ you're trying to create.

2nd edit: So regarding 3-manifolds I think your question has something of an answer now, right? $pi_n M$ is $pi_n$ of the universal cover provided $n > 1$. The universal cover of a geometric 3-manfold is homeomorphic to $mathbb R^3$ or $S^3$. So by climbing up the JSJ and connect sum decomposition of a 3-manifold, the universal cover is diffeomorphic to a punctured $S^3$ -- the number of punctures is either $0$, $1$, $2$ or a Cantor set's worth of punctures. In the Cantor set case we're giving this complement the compactly generated topology induced from the Cantor set complement's subspace topology. So among other things, $pi_2 M$ is a direct sum of copies of $mathbb Z$, similarly $pi_3 M$, torsion first appears in $pi_4 M$. The complement you think of as a directed system of wedges of $S^2$'s so the Hilton-Milnor theorem tells you the homotopy groups.