Elaborating on Alex Suciu's first comment, let me offer a second way of computing $\pi_3(M\#N)$ which works for any pair of $3$-manifolds with non-trivial fundamental groups (i.e. none of them is $S^3$, which is uninteresting since it is a unit for $\#$).

For any (CW-)space $X$ with universal cover $\tilde X$ there is an exact sequence (due to JHC Whitehead)

$$H_4(\tilde X)\rightarrow\Gamma \pi_2(X)\rightarrow\pi_3(X)\rightarrow H_3(\tilde X)\rightarrow 0.$$

Here $\Gamma$ is Whitehead's universal quadratic functor (see below). If $X$ is a $3$-manifold with inifinite fundamental group then $\tilde X$ is an open $3$-manifold and $H_4(\tilde X)=H_3(\tilde X)=0$, hence
$$\pi_3(X)=\Gamma \pi_2(X).$$
This applies to $X=M\#N$ since $\pi_1(X)=\pi_1(M)*\pi_1(N)$ is infinite.

A map between abelian groups $\gamma\colon A\rightarrow B$ is *quadratic* if $\gamma(a)=\gamma(-a)$ and the map $A\times A\rightarrow B\colon (a_1,a_2)\mapsto\gamma(a_1+a_2)-\gamma(a_1)-\gamma(a_2)$ is bilinear. The abelian group $\Gamma A$ is characterized as the target of the universal quadratic map $A\rightarrow \Gamma A$. If $A$ is free abelian with basis $B\subset A$ then $\Gamma A$ is free abelian with basis
$$\{\gamma(b)\,;\,b\in B\}\cup \{\gamma(b_1+b_2)-\gamma(b_1)-\gamma(b_2)\,;\,b_1<b_2\in B\}.$$
Here $<$ is any total order in $B$. In our case $X=M\#N$, $\pi_2(X)=\mathbb Z[\pi_1(X)]$ is free abelian, as pointed out by Alex Suciu.

The isomorphism $\pi_3(X)=\Gamma \pi_2(X)$ holds since precomposition with the Hopf map $\eta\colon S^3\rightarrow S^2$ is the universal quadratic map $\eta^*\colon\pi_2(X)\rightarrow\pi_3(X)$, $\eta^*(f)= f\circ\eta$, in this case. Notice that this is the same computation as in Allen Hatcher's excellent answer since it's known that $\eta^*(f+g)-\eta^*(f)-\eta^*(f)=[f,g]$ coincides with the Whitehead product.