How to prove the connected sum of two closed aspherical n-manfolds (n >2) is not asperical?

If $N$ is an open contractible manifold of dimension at least two, then it has one end. For instance, you can compute the zeroth reduced homology of $N$ minus a closed ball $B$ by using the long exact sequence for $(N,B)$ by using excision and the fact that $N$ is contractible.

If $M$ is a closed aspherical manifold of dimension at least three and $M$ is a connected sum of $M_1$ and $M_2$, then the fundamental group of $M$ is the free product of the the fundamental groups of $M_1$ and $M_2$.

If $M_1$ and $M_2$ are closed aspherical manifolds of dimension at least three, then their fundamental groups are nontrivial, and so $\pi_{1}(M)$ would be a nontrivial free product. But the universal universal cover of $M$ is one ended, and one ended groups do not decompose as nontrivial free products.

Here's a different way to see it. Let $M$ and $N$ be aspherical of dimension at least 3. Then the wedge $M \vee N$ is aspherical (but not a manifold). Let $M\sharp N$ be the connected sum. Then we get a collapse map $M\sharp N \to M \vee N$ given by pinching to a point the embedded $(n-1)$-sphere you used to form the connected sum. It is trivial to check that this map is $(n-1)$-connected, and therefore an isomorphism of fundamental groups.

If the connected sum were acyclic aspherical then it would have to be homotopy equivalent to the wedge, since the latter is acyclic aspherical and has the same fundamental group. Moreover, the collapse map would have to give such a homotopy equivalence. But this is ludicrous, since that would violate the long exact homology sequence of the cofibration $$ S^{n-1} \to M\sharp N \to M \vee N . $$

In fact if $M$ is a closed aspherical $n$-manifold, then $\pi_1(M)$ does not split as an amalgamated product or HNN extension over a subgroup $H$ of cohomological dimension $\le n-2$. For concreteness suppose $\pi_1(M)=A*_C B$. Look at Mayer-Vietoris sequences of the splitting in dimension $n$ with $\mathbb Z_2$-coefficients. The factors $A$, $B$ have infinite index in $\pi_1(M)$, so their $n$-dimensional cohomology vanish, and so does the $(n-1)$-cohomology of $C$. By exactness, $H^n(M;\mathbb Z_2)$ vanishes, so $M$ cannot be closed aspherical.