The answer is that this never happens for manifolds which are of finite type in the sense that they are homotopy equivalent to finite CW complexes. Serre showed that a simply connected finite CW complex has infinitely many nonzero homotopy groups.

Loosely, there's a kind of uncertainty principle relating homotopy and cohomology: it's hard for a space to simultaneously have few homotopy groups and few cohomology groups. So on the one hand it's hard for classifying spaces $B^n A$ to have bounded cohomology (I think they never have bounded cohomology if $n$ is even?), and on the other hand it's hard for finite CW complexes to have bounded homotopy.

The cohomology of classifying spaces $B^n A$ is extensively studied because they describe cohomology operations, so presumably someone who's more familiar with these can tell you more about no-go results from this direction. It's not hard to show that $B^n \mathbb{Z}$ has nonzero cohomology in arbitrarily high degrees when $n$ is even; you can take the cohomology operations to be cup powers. Similarly it's not hard to show that $B^n \mathbb{Z}_2$ has nonzero cohomology in arbitrarily high degrees for all $n$; here you can also take cup powers, but Steenrod operations are also available.