Fix $G$, a finitely ~~generated~~ presented group.

It is known that for every $k > 3$ there is a closed $k$-manifold whose fundamental group is $G$. Similarly, there is a topological space with fundamental group $G$ and all higher homotopy groups trivial.

However, even for simple examples such as when $G \cong \mathbf{Z}_2$, such a topological space is not a manifold. It seems like the problem with these spaces really lies in the infinite constructions process adding in cells of arbitrarily high dimension. So instead if we only require the first $n$ homotopy groups to be trivial can we still work with manifolds. That is,

Is it true that for each $n > 1$ there is a closed manifold $M$ such that $\pi_1(M) \cong G$ and for $1 < i \leq n$, $\pi_i(M)$ is trivial?

Note that if we allow $M$ to be a non-compact manifold / a manifold with boundary then the answer is yes. This follows as we can always find a finite simplicial complex $X$ whose fundamental group is $G$. By correctly adding $i$-cells (for $1 < i \leq n$) we obtain a simplical complex $X'$ with $\pi_1(X') \cong G$ and for $1 < i \leq n$, $\pi_i(M)$ trivial. By embedding $X'$ in a suitably high dimensional Euclidean space and taking an closed / open regular neighbourhood we obtain $M$, a non-compact manifold / manifold with boundary with the required properties.

Assuming that the answer to the first question is yes, can we also get manifolds of almost any dimension that we like?