$\mathbb{Q}(\sqrt{p+\sqrt{p}})$ is Galois over $\mathbb{Q} \iff p=n^2+1$ for a prime $p$.

Suppose $p$ is prime. The Galois closure of $L=\Bbb Q(\sqrt{p+\sqrt p})$ is $\Bbb Q(\sqrt{p+\sqrt p},\sqrt{p-\sqrt p})$. So $L$ is Galois over $\Bbb Q$ iff $p-\sqrt p$ is a square in $L$.

By Kummer theory, $p-\sqrt p$ is a square in $L$ iff either it is a square in $K=\Bbb Q(\sqrt p)$ or if $(p+\sqrt p)(p-\sqrt p)$ is a square in $K$. In the former case $N(p-\sqrt p)=p^2-p$ must be a square in $\Bbb Q$, which it isn't as it's a multiple of $p$ but not of $p^2$. In the latter case $(p+\sqrt p)(p-\sqrt p)=p(p-1)$ is a square in $K$. The rationals that are squares in $K$ are the squares of rationals or $p$ times the squares of rationals. $p(p-1)$ is not a square of a rational, and is $p$ times a square of a rational iff $p-1=n^2$ where $n$ is an integer.