3-manifold examples of homomorphisms between fundamental groups that are not induced by continuous maps

To piggyback on Moishe's answer: Every map $T^3 \to T^3 \# T^3$ has degree zero. Thus one sees immediately by investigating the cohomology that the induced map on $H^1$ is not surjective, and in particular there are many homomorphisms that cannot be realized.

Suppose the map has nonzero degree; make it transverse to the connected sum sphere. Write $T^3 = M_1 \cup M_2$, where $M_i$ is the inverse image of the $i$th component of the connected sum; the boundary $\partial M_1 = \partial M_2 = f^{-1}(S^2)$. Then we have a map $(M_i, \Sigma) \to (T^3 \setminus D^3, S^2)$. Then the map $H_3(M_i, \Sigma) \to H_3(T^3 \setminus D^3, S^2)$ is still multiplication-by-$d$ (degree is an entirely local property!) Collapsing the 2-sphere boundary to get a map $(M_i, \Sigma) \to (T^3, *)$, this is again degree $d$. This implies that $\Sigma$ is not a 2-sphere, since $T^3$ is irreducible, and any map $S^3 \to T^3$ has degree zero.

I claim that this implies $\Sigma$ is incompressible inside the domain. For suppose you had a compressing disc. Because $\pi_2T^3$ is trivial, this implies the compressing disc is homotopic to a disc inside $\Sigma$; but by assumption the original loop was not null in $\Sigma$. This gives a contradiction.

Now an incompressible surface is $\pi_1$-injective by the loop theorem. So it must be a map $T^2 \to T^3$ that's $\pi_1$-injective; one can verify that this means that the surface is homologically nontrivial in $T^3$. But it bounds the $M_i$! So we've arrived at a contradiction to the original claim, and thus every map $T^3 \to T^3 \# T^3$ is degree zero.

This argument is not particularly special to $T^3$; you should be able to mimic it to get many more examples. The general result is that a map $Y \to N$ has degree zero if $N$ has more prime summands than $Y$ has disjoint, non-parallel, incompressible surfaces.

This is a partial answer to your question, based on this mathoverflow thread. It was observed in the mathoverflow answer that the first obstruction for existence of a continuous map $X\to Y$ (inducing the given homomorphism of fundamental groups) is in $H^3(X, \pi_2(Y))$. In the case when $Y$ is a lens space, $\pi_2(Y)=0$, hence the first obstruction vanishes. If $X$ is 3-dimensional (as in your question), there are no further obstructions. This answers the question about the case of lens spaces.

Thus, in order to find examples when a map between 3-manifolds does not exist you need to consider target manifolds which are nontrivial connected sums (as they have to have $\pi_2\ne 0$). That I am not sure how to do. It seems reasonable to consider maps to $T^3\# T^3$ as a test case.