Finding the intermediate fields of $\Bbb{Q}(\zeta_7)$.

Here's a trick that often works in this kind of problem.

Take an element $\alpha$ of $F$ --- an element that generates $F$ over the ground field.

Take a subgroup $H$ of the Galois group.

Form $\sum_{\sigma{\rm\ in\ }H}\sigma(\alpha)$. This is guaranteed to be invariant under $H$ and thus to live in some subfield. Unless you're unlucky, it generates the fixed field of $H$.

In particular, this is what @ccorn has done in the other answer.

There is no reason to assume $\mathrm{i}\in F$, so try another $\beta$. I would suggest $$\beta = \zeta_7 + \zeta_7^2 + \zeta_7^4$$ You will easily verify that $\beta+\bar{\beta}=-1$ and $\beta\bar\beta=2$ (thus $\beta$ is irrational) and that $$\mathbb{Q}[X]\ni\operatorname{minpoly}_{\mathbb{Q}}(\beta)=X^2-(\beta+\bar{\beta})X+(\beta\bar{\beta}) = X^2+X+2$$ Thus $[\mathbb{Q}(\beta):\mathbb{Q}] = 2$.

Edit: Fixed the sloppy notation for $\operatorname{minpoly}_{\mathbb{Q}}(\beta)$. See Gerry Myerson's answer for finding a candidate $\beta$ in general.

Edit 2: $\mathrm{i}\not\in F$ because $\mathrm{i}\zeta_7$ is a primitive $28$th root of unity, so $F(i)$ contains $\zeta_{28}$, whereas $[\mathbb{Q}(\zeta_{28}):\mathbb{Q}] = 12 \neq 6 = [F:\mathbb{Q}]$.

Edit 3: Clarified way to the minimal polynomial.