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.