Second Homotopy Group of Graph Manifolds A graph manifold is a closed 3-manifold $M$ that admits a finite collection of disjoint embedded tori $\mathcal{T}$ so that $M \setminus \mathcal{T}$ is a disjoint union of Seifert fibred spaces (i.e. spaces admitting a foliation with a circle as fibre).
My interest in this class of manifolds lies in its characterization through the methods of geometrization: A 3-manifold is a graph manifold if and only if its simplicial norm vanishes.
Question: What is known about the second homotopy group of graph manifolds? I am particulary interested in an answer to the following question: How many graph manifolds have finite second homotopy group?
The only hint I could find was the following result due to D. Hume (cf. http://arxiv.org/abs/1112.0263v2): The universal cover of any graph manifold quasi-isometrically embeds in the product of three metric trees.
 A: Have you thought of using the 2-dimensional Seifert-van Kampen theorem?  
R. Brown and P.J. Higgins, ``On the connection between the second
relative homotopy groups of some related spaces'', Proc.
London Math.  Soc. (3) 36 (1978) 193-212.
This gives information on the second relative homotopy group $\pi_2(X_2,X_1,x)$ as a crossed module over $\pi_1(X_1,x)$,  in the case $X$ is a union of subspaces $U^a,a \in A$, and  in terms of the relative homotopy group  $\pi_2(X_2 \cap U^a, X_1\cap U^a)$ as a crossed module over $\pi_1(X_1\cap U^a,x)$ under certain connectivity assumptions. Further details, including more modes  of computation,  are in Part I of the book advertised on 
http://pages.bangor.ac.uk/~mas010/nonab-a-t.html
I put forward this as an idea, without having done any work on the problem posed. However, since this 2-d SvKT seems little known, it seems worth putting forward the possibility. Note the point that if the theorem applies, then it determines the 2-type itself as a crossed module, but it can still be a problem to compute the second homotopy group. Some methods are available, but few have worked on the area. 
If this idea is inapplicable to this case, I won't be worried. But the question asked is about manifolds which are unions in a prescribed way, so it might work. 
A short list of papers with applications of such higher homotopy Seifert-van Kampen theorems is on 
http://pages.bangor.ac.uk/~mas010/pdffiles/appHvKT.pdf
A: I thought perhaps I should write up my comments to Agol's answer as a separate answer itself.
Proving asphericity of a graph manifold $M$ with $\pi_1$-injective tori can be done from the point of view of rather easy homotopy theory, using the Scott-Wall concept of graphs of spaces. By splitting along the torus decomposition of $M$, one gets a finite graph of spaces whose vertex spaces are Seifert fibered 3-manifolds and whose edge spaces are toruses, both of which are $K(\pi,1)$'s. Assuming that each inclusion of a torus edge space into an incident Seifert Fibered vertex space is $\pi_1$-injective ---- which as Agol says is always true when the Seifert fibered spaces are not solid tori --- the universal covering of the graph of spaces is a tree of contractible spaces. More precisely, it is a tree of spaces over the Bass-Serre tree $T$ of the splitting, the vertex spaces upstairs are universal covers of the Seifert fibered spaces which are contractible, and the edge spaces upstairs are universal covers of the torusses which are contractible. 
Any tree of contractible spaces is contractible, and therefore the downstairs space (in this case the graph manifold) is aspherical. Contractibility of a tree of contractible spaces can be checked by mapping a sphere in, and applying induction on the (finite) number of edges of the subtree $\tau \subset T$ whose edges spaces intersect the image of the sphere. The induction step is to homotope the map of the sphere to push it off an edge space incident to a valence 1 vertex space of $\tau$, reducing the number of edges of $\tau$ by $1$. In the base step, where $\tau$ has one edge, the sphere map is homotopic to a constant within an edge space.
A: The answer to your question is almost trivial: if the tori/Klein bottles are required to be $\pi_1$-injective, then the manifold will be a $K(\pi,1)$, so $\pi_2$ will vanish. Also, notice in the case of Seifert-fibered spaces, the universal cover is either $\mathbb{R}^3$ or $S^3$, and thus again $\pi_2=0$.
However, there is a bit of discrepancy about what is considered a graph manifold in the literature. For 3-manifold topologists, the manifolds are required to be irreducible (so the tori are $\pi_1$-injective), and thus $\pi_2=0$. However, for geometers, sometimes they allow some of the tori to be compressible, so one obtains solid tori for some of the Seifert pieces. This is useful, because such graph manifolds have $F$-structures, and thus can admit sequences of metrics which collapse in the sense of Cheeger-Gromov. So if you're asking about this flavor of graph manifolds, then the answer is that $\pi_2$ is infinitely generated, unless you happen to have something finitely covered by $S^1\times S^2$, in which case it's $\mathbb{Z}$, or $S^3$ or $R^3$, in which case $\pi_2$ is trivial. In particular, $\pi_2$ is never finite. This kind of graph manifold opens up a can of worms since it is highly non-canonical: there are graph manifold structures on $S^3$ associated to any iterated cable link. See these slides for the classification of universal covers of closed 3-manifolds. 
