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 $N$? 



How to compute $\pi_{3}$ of $L(p,q)\# L(p',q')$?
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.


