= Solution
For $n=p$ prime, $[K:\mathbb Q]=p-1$ and the cyclic group $\operatorname{Gal}(F/K)$ has order either $1$ or $p$. Therefore
$$
\boxed{[F:\mathbb Q]\in\{p-1,\ p(p-1)\}.}
$$
These correspond respectively to $a$ being or not being a $p$th power in $K$. In fact it is a $p$th power in $K$ precisely when it is a $p$th power in $\mathbb Q$. To see the nontrivial implication, suppose $\alpha\in K$ and $\alpha^p=a$. Taking the <field norm> gives $b^p=a^{p-1}$ with $b=N_{K/\mathbb Q}(\alpha)\in\mathbb Q^\times$. Then $(a/b)^p=a$, so $a$ is already a rational $p$th power.
Both degrees occur: $a=1$ gives $p-1$, while $a=2$ is not a rational $p$th power and gives $p(p-1)$. This includes $p=2$, with possible degrees $1$ and $2$.
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