Is there a less ad hoc way to find the degree of this extension? I want to find the degree of $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}$. I observe that
$3+2\sqrt{2}=2+2\sqrt{2}+1=(\sqrt{2}+1)^2$ so
$$
\mathbb{Q}(\sqrt{3+2\sqrt{2}})=\mathbb{Q}(\sqrt{2}+1)=\mathbb{Q}(\sqrt{2})
$$
so the degree is 2.
Is there a more mechanical way to show this without noticing the factorization?

This reduces to checking if the radical $\:\sqrt{3+2\sqrt{2}}\:$ denests. While there are general algorithms, simple cases like this can be tackled by employing an easy formula that I discovered as a teenager.
Simple Denesting Rule $\rm\quad \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \  \color{brown}{divide\ out}\ \sqrt{trace} $
E.g. $\ 3 + 2\sqrt{2}\ $ has norm $\ (3+2\sqrt{2})\:(3-2\sqrt{2})\ =\ 9 - 4\cdot 2\ =\ 1$
So $\rm\:\color{blue}{subtracting\ out}\ \sqrt{norm}\ = \pm1\ $ yields $\rm\ 3 + 2\sqrt{2}\: -\: \pm1\  =\  2 + 2\sqrt{2}\ \ or\ \ 4 + 2\sqrt{2} $
$\rm\qquad \sqrt{trace(2+2\sqrt{2})}\ =\ \sqrt{4}\ =\ 2,\quad\ \ so\ \quad\ \color{brown}{dividing\ it\ out}\ \ \ (2+2\sqrt{2})/2\ =\  1+\sqrt{2}$
$\rm\qquad \sqrt{trace(4+2\sqrt{2})}\ =\ \sqrt{8}\ =\  2\sqrt{2}\quad so\quad \color{brown}{dividing\ it\ out}\ \ \ (4+2\sqrt{2})/(2\sqrt{2})\: =\: \sqrt{2}+1$
Indeed $\rm\ (1 + \sqrt{2})^2\ =\ 3 + 2\sqrt{2}\:.\ $ It is an easy exercise to check that the formula is correct.
With experience from a few worked examples, one can swiftly mentally denest such radicals.
