Order of torsion group What can one say about the order of the torsion group of an elliptic curve defined over the compositum of all quadratic extensions of $\mathbb{Q}$ ?
 A: I think the papers you should have a look at are those two by Fujita, great material and very well-written (in my humble opinion):
1) Y. Fujita, Torsion subgroups of elliptic curves with non-cyclic torsion over ${\mathbb Q}$ in elementary abelian 2-extensions of ${\mathbb Q}$, Acta Arith. 115 (2004) 29–45. MR2102804 (2005j:11041)
2) Fujita, Y.: Torsion subgroups of elliptic curves in elementary abelian $2$-extensions of ${\mathbb Q}$. J. Number Theory, 114, 124–134 (2005) MR2163908 (2006h:11055)
Hope this helps.
A: Mazur's theorem is not needed here, and besides, it applies only to elliptic curves themselves defined over $\mathbb{Q}$ rather than the said compositum $K^{(2)}$ of all quadratic extensions of $\mathbb{Q}$ (which is how I understand the question).
A simple proof is available by purely local methods. First, if $F/\mathbb{Q}_p$ is a finite extension, then any abelian variety $A/F$ has $|A(F)_{\mathrm{tors}}| < \infty$. This follows from the theory of the formal group; for the elliptic curve case, see Chapter IV, Theorem 6.4 in Silverman's book (supplemented by Ch. VII Ex. 7.6 as noted by  René).
Given this, fix any prime. Since $\mathbb{Q}_p$ has just finitely many quadratic extensions, it is possible to embed $K^{(2)}$ in a finite extension $F/\mathbb{Q}_p$. Then the stated finiteness follows from the above quoted result.
A: Extended comment.  OP Asked if there was an easy way to see that the group is finite.  Indeed, if $E$ is an elliptic curve over $\mathbb{Q}$ and $K$ is a Galois extension of $\mathbb{Q}$ with only finitely many roots of unity, then $E(K)_\text{tors}$ is finite.  You can't have full $n$-torsion if there is not a primitive $n^{\text{th}}$ root of unity in $K$, thanks to the Weil pairing.  If you have "half" $n$-torsion, then $\operatorname{Gal}(K/\mathbb{Q})$ acts on this cyclic group of order $n$, which means there is an $n$-isogeny which is defined over $\mathbb{Q}$.  For large enough $n$, this is impossible, by Mazur.  (I learned this argument from Filip Najman.)
(And it's easy to see that the compositum of all quadratic fields has only finitely many roots of unity.  In fact, you could count them on your fingers.)
EDIT: You may need to get creative, as comments suggest, to accomplish this feat with your fingers.
