Somehow, the theme of symmetrization often doesn't come across very clearly in many expositions of Galois theory. Here is a basic definition:
Definition. Let $F$ be a field, and let $G$ be a finite group of automorphisms of $F$. The symmetrization function $\phi_G\colon F\to F$ associated to $G$ is defined by the formula
$$
\phi_G(x) \;=\; \sum_{g\in G} g(x).
$$
Example. Let $\mathbb{C}$ be the field of complex numbers, and let $G\leq \mathrm{Aut}(\mathbb{C})$ be the group $\{\mathrm{id},c\}$, where $\mathrm{id}$ is the identity automorphism, and $c$ is complex conjugation. Then $\phi_G\colon\mathbb{C}\to\mathbb{C}$ is defined by the formula
$$
\phi_G(z) \;=\; \mathrm{id}(z) + c(z) \;=\; z+\overline{z} \;=\; 2\,\mathrm{Re}(z).
$$
Note that the image of $\phi$ is the field of real numbers, which is precisely the fixed field of $G$. This example generalizes:
Theorem. Let $F$ be a field, let $G$ be a finite group of automorphisms of $F$, and let $\phi_G\colon F\to F$ be the associated symmetrization function. Then the image of $\phi_G$ is contained in the fixed field $F^G$. Moreover, if $F$ has characteristic zero, then $\mathrm{im}(\phi_G) = F^G$.
Of course, since $\phi_G$ isn't a homomorphism, it's not always obvious how to compute a nice set of generators for its image. However, in small examples the goal is usually just to produce a few elements of $F^G$, and then prove that they generate.
Let's apply symmetrization to the present example. You are interested in the field $\mathbb{Q}(\zeta_7)$, whose Galois group is cyclic of order $6$. There are two subgroups of the Galois group to consider:
The subgroup of order two: This is the group $\{\mathrm{id},c\}$, where $c$ is complex conjugation. You have already used your intuition to guess that $\mathbb{Q}(\zeta_7+\zeta_7^{-1})$ is the corresponding fixed field. The basic reason that this works is that $\zeta_7+\zeta_7^{-1}$ is the symmetrization of $\zeta_7$ with respect to this group.
The subgroup of order three: This is the group $\{\mathrm{id},\alpha,\alpha^2\}$, where $\alpha\colon\mathbb{Q}(\zeta_7)\to\mathbb{Q}(\zeta_7)$ is the automorphism defined by $\alpha(\zeta_7) = \zeta_7^2$. (Note that this indeed has order three, since $\alpha^3(\zeta_7) = \zeta_7^8 = \zeta_7$.) The resulting symmetrization of $\zeta_7$ is
$$
\mathrm{id}(\zeta_7) + \alpha(\zeta_7) + \alpha^2(\zeta_7) \;=\; \zeta_7 + \zeta_7^2 + \zeta_7^4.
$$
Therefore, the corresponding fixed field is presumably $\mathbb{Q}(\zeta_7 + \zeta_7^2 + \zeta_7^4)$.
All that remains is to find the minimal polynomials of $\zeta_7+\zeta_7^{-1}$ and $\zeta_7 + \zeta_7^2 + \zeta_7^4$. This is just a matter of computing powers until we find some that are linearly dependent. Using the basis $\{1,\zeta_7,\zeta_7^2,\zeta_7^3,\zeta_7^4,\zeta_7^5\}$, we have
$$
\begin{align*}
\zeta_7 + \zeta_7^{-1} \;&=\; -1 - \zeta_7^2 - \zeta_7^3 - \zeta_7^4 - \zeta_7^5 \\
(\zeta_7 + \zeta_7^{-1})^2 \;&=\; 2 + \zeta_7^2 + \zeta_7^5 \\
(\zeta_7 + \zeta_7^{-1})^3 \;&=\; -3 - 3\zeta_7^2 - 2\zeta_7^3 - 2\zeta_7^4 - 3\zeta_7^5
\end{align*}
$$
In particular, $(\zeta_7+\zeta_7^{-1})^3 + (\zeta_7+\zeta_7^{-1})^2 - 2(\zeta_7+\zeta_7^{-1}) - 1 = 0$, so the minimal polynomial for $\zeta_7+\zeta_7^{-1}$ is $x^3 + x^2 - 2x - 1$. Similarly, we find that
$$
(\zeta_7 + \zeta_7^2 + \zeta_7^4)^2 \;=\; -2 - \zeta_7 - \zeta_7^2 - \zeta_7^4
$$
so the minimal polynomial for $\zeta_7 + \zeta_7^2 + \zeta_7^4$ is $x^2+x+2$.
There is a general method to deal with these sorts of extensions. Suppose we have $F$ a field of characteristic not equal to $2$ and suppose we adjoin $\alpha = \sqrt{a + b\sqrt{c}}$ Suppose that $b\neq0$, $a^2 - b^2c = h^2$ for some $h\in F$. Let $f(x)$ be the minimal polynomial of $\sqrt{a + b\sqrt{c}}$ (put some nice conditions on $a,b,c$ so that $F(\alpha)$ is an extension of degree $4$).
Let us call $\alpha_1 = \alpha$ , $\alpha_2 = \sqrt{a - b\sqrt{c}}$, $-\alpha_1= \alpha_3$ and $-\alpha_2 = \alpha_4$. We choose the following presentation for $V_4$ (the Klein 4- group isomorphic to $\Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$): $$V_4 = \{e, (12)(34), (13)(24),(14)(23)\}.$$ Note that $(14)(23) = \big((13)(24)\big)\big((12)(34)\big)$. Then we can see that each cycle in $V_4$ acts on the roots according to the numbering convention: For example the cycle $(12)(34)$ sends $\alpha_1 \mapsto \alpha_2$, $\alpha_3\mapsto \alpha_4$. Now for the sake of simplicity let us call
$$\theta = (12)(34), \hspace{5mm} \gamma = (13)(24), \hspace{5mm} \gamma\theta = (14)(23).$$
Then it is easy to see that $\theta(\alpha_1 + \alpha_2) = \alpha_1 + \alpha_2, \gamma\theta(\alpha_1 - \alpha_2) = \alpha_1 - \alpha_2$. Now because $\gamma(\alpha_1) = \alpha_3$, squaring both sides we see that $\gamma(\sqrt{c} ) = \sqrt{c}$. It follows from these observations that
$$F(\alpha_1 + \alpha_2) \subset E^{\langle \theta \rangle}, \hspace{5mm} F(\sqrt{c}) \subset E^{\langle \gamma \rangle}, \hspace{5mm} F(\alpha_1 - \alpha_2) \subset E^{\langle \gamma\theta\rangle }.$$
By computing minimal polynomials/degrees we get that these subset inclusions are actually equalities and so you are done.
Edit: As Jyrki points out, the only presentation for $V_4$ that we can choose is the one given in my answer as that presentation is the only one which gives a transitive subgroup of $S_4$.
Best Answer
For an involutive automorphism $\phi$, for all elements $s$, we have that $s+\phi(s)$ and $s\cdot \phi(s)$ are always fixed, so they have a chance to generate the subfield..
Update: interestingly, $$ \begin{align} \left(\sqrt{5+\sqrt{21}} + \sqrt{5-\sqrt{21}}\right)^2 &= 14 \\ \left(\sqrt{5+\sqrt{21}} - \sqrt{5-\sqrt{21}}\right)^2 &= 6. \end{align} $$ Hmm.. In other words, the following 3 intermediate fields we have: $$\Bbb Q(\sqrt 6),\ \Bbb Q(\sqrt{14}),\ \Bbb Q(\sqrt{21})$$