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 $\mathbb{Z}$. \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.
