The intuitive idea is that the sphere connected the two manifolds is not contractible, which implies the (n1)th homotopy group is not zero. Another argument, which I am not totally understand, uses the fact that the universal covering of an aspherical manifold has only one end. I am wondering if someone here could clarify these for me or give me a new argument.
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 $(n1)$sphere you used to form the connected sum. It is trivial to check that this map is $(n1)$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
$$
S^{n1} \to M\sharp N \to M \vee N .
$$

$\begingroup$ By the way, this argument generalizes to connected sums of acyclic Poincare duality spaces. $\endgroup$ – John Klein Jan 13 '11 at 4:17

$\begingroup$ In the second paragraph, should 'acyclic' be 'aspherical'? $\endgroup$ – Michael Albanese Jul 3 at 17:13

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 $\le n2$. For concreteness suppose $\pi_1(M)=A*_C B$. Look at MayerVietoris 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 $(n1)$cohomology of $C$. By exactness, $H^n(M;\mathbb Z_2)$ vanishes, so $M$ cannot be closed aspherical.
Darryl McCullough gives a very complete answer to your question in his "Connected Sums of Aspherical Manifolds" paper. Let $M$ be the connected sum of $g$ aspherical manifolds of dimension $d\geq 3$. Its universal cover $\tilde M$ is homotopy equivalent to $\bigvee_{\rho \in \pi_1(M)} \bigvee_{i=1}^{g1}S^{d1}$ (and in particular is not contractible).
I wonder if there is a short proof of this along the lines of Robert Bell's answer.