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 typo in the statement of Lemma 67 in your source. The $n$th roots of unity are in $\Bbb{F}_p$ only if $n\mid p-1$ or, iff $p\equiv1\pmod n$ (not $n\equiv1\pmod p$ as is written there). Therefore that Lemma does not apply.
In fact, the polynomial $x^{15}-2$ is NOT irreducible in $\Bbb{F}_7[x]$. This follows trivially from the fact that $3^5=243\equiv-2\pmod 7$. Therefore
$$
x^{15}-2=(x^3)^5+3^5=(x^3+3)(x^{12}-3x^9+3^2x^6-3^3x^3+3^4).
$$
We immediately see that $x^3+3$ has no zeros in $\Bbb{F}_7$ (the cubes in that field are $0,\pm1$), so it is irreducible. Therefore the polynomial has a zero $\alpha$ in $\Bbb{F}_{7^3}$.
To get the splitting field of $x^{15}-2$ we need, as you observed, the primitive 15th roots of unity. We easily see that
$$
7^4=2401\equiv1\pmod{15}.
$$
The multiplicative group of the field $\Bbb{F}_{7^4}$ is cyclic of order $7^4-1$, and thus it contains a primitive 15th root of unity $\zeta$.
A consequence of all this is that the splitting field of this polynomial is
$$
\Bbb{F}_7[\alpha,\zeta]=\Bbb{F}_{7^{12}}.
$$
Best Answer
Since we want the degree of an irreducible factor to be equal to one, we want $$ \text{order} (2^k) =1 $$ in $(\mathbb{Z} / 5\mathbb{Z})^\ast$. The only element with this order is 1. Therefore we search the smallest $k$ such that $2^k\equiv 1\mod 5$. A bit puzzzling gives us $$ 2^1=2\\ 2^2=4\\ 2^3=8=3\\ 2^4=16=1. $$ Therefore the splitting field of $\Phi_5$ should be $\mathbb{F}_{2^4}$.
Is this correct?