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> For a fixed $d$, is there a relationship between the homotopy groups of smooth $d$-manifolds?
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The $d=1$ case is trivial, but I already don't know how to approach $d=3$ (I should have said that the case of $d=2$ is simple as well, since there is only a sphere to consider, but I don't know how to formulate the property of "having the same homotopy groups as $S^2$" in a simpler way).


Note about the discussion on the comments: it's unreasonable to expect an easy *complete* characterization of homotopy groups of $S^2$, even less for other manifolds. But I think one could try some *partial* relations. An interesting relationship would be: for some $d$, the groups $\pi\_n$ can be determined from groups $\pi\_m$ for $m<N<n$ (this is unlikely to be true though).



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. 


