Let $L(p,q)$ be a 3-dimensional lens space, and let $L(p',q')$ be another. Is there any known result concerning the 3rd homotopy group of the connected sum $L(p,q)\# L(p',q')$? If not, I am interested in computing it, but I have no idea.
More generally, can we find a method to compute $\pi_{3}(M\# N)$ for general prime 3-manifolds $M$ and $N$?
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.
One can compute $\pi_3$ completely by using the fact that it is isomorphic to $\pi_3$ of the universal cover. For a connected sum $M=P\# Q$ of lens spaces $P$ and $Q$ one obtains the universal cover $\widetilde M$ in the following way. Let $P'$ and $Q'$ be obtained from $P$ and $Q$ by deleting an open ball from each. Their universal covers are $S^3$ with a finite number of disjoint balls removed, and $\widetilde M$ is obtained by gluing together infinitely many copies of the universal covers of $P'$ and $Q'$ along their boundary spheres in a treelike pattern, so that the result is simply-connected. A 3-sphere with finitely many disjoint balls deleted is homotopy equivalent to a wedge of 2-spheres, with the number of 2-spheres being one less than the number of deleted balls. It's not too hard then to figure out that $\widetilde M$ is homotopy equivalent to the wedge of an infinite number of 2-spheres (corresponding to a subset of the gluing spheres). Computing $\pi_3$ of a wedge of 2-spheres is then not too difficult: it is free abelian with basis given by the Hopf maps $S^3\to S^2$ for the individual $S^2$ wedge summands and the Whitehead products $S^3\to S^2\vee S^2$ for each pair of distinct 2-spheres in the wedge sum. (A textbook reference for this is Example 4.52 of my algebraic topology book.)
Incidentally this sort of calculation works more generally for $\pi_3$ of any noncompact simply-connected 3-manifold. Such a manifold has the homotopy type of a 2-complex, and since it's simply-connected it must be homotopy equivalent to a wedge of 2-spheres.
