I agree with your choice of splitting field, although I think your solution is probably a bit longer than is standard. Just note that, if $E$ is the splitting field of $f$ over $\mathbb{Q}$, then $\sqrt{2}, \sqrt{3}, \sqrt{5} \in E$, so $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}) \subseteq E$. In fact $f$ splits in $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$, which means that $E = \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$.
Now, $E$ is a splitting field over $\mathbb{Q}$, which means that (by a general result of Galois Theory) it is a normal extension of $\mathbb{Q}$. Furthermore, $\mathbb{Q}$ is a field of characteristic zero, which means that any finite field extension of $\mathbb{Q}$ is separable (fields of characteristic zero are perfect), so $E/\mathbb Q$ is a normal, separable extension, hence it is a Galois extension.
Since $E/\mathbb{Q}$ is a finite Galois extension, its Galois group $G = \text{Gal}(E/\mathbb{Q})$ has $\lvert G \rvert = [E:\mathbb Q]$, and by the linked question, $[E:\mathbb{Q}] = 8$, so $G$ is a group of order $8$. This means that there are eight $\mathbb{Q}$-automorphisms of $E$. Let $\varphi:E\to E$ be such a $\mathbb{Q}$-automorphism. For each of $d = 2,3,5$, we have $\varphi(\sqrt{d})^2 = \varphi(d) = d$, which means that $\varphi(\sqrt{d}) = \pm \sqrt{d}$. There are thus $2^3 = 8$ choices for the images of the $\sqrt{d}$ under $\varphi$. Note that each such choice determines $\varphi$ fully because $E$ is generated over $\mathbb{Q}$ by these $\sqrt{d}$. Since there are eight such choices, and we know that $\lvert G \rvert = 8$, so in fact the elements of $G$ are precisely the $\mathbb{Q}$-automorphisms induced by
$$
(\sqrt{2},\sqrt{3},\sqrt{5}) \mapsto (\epsilon_1\sqrt{2},\epsilon_2\sqrt{3},\epsilon_3\sqrt{5})
$$
for all possible choices of $\epsilon_i =\pm 1$.
If we identify the roots of $f$ with the set $\Omega = \{1,2,3,4,5,6\}$ via
$$
(\sqrt{2}, -\sqrt{2},\sqrt{3},-\sqrt{3},\sqrt{5},-\sqrt{5}) \leftrightarrow (1,2,3,4,5,6)
$$
then $G$ may be viewed as a subgroup of the symmetric group $S_6$ generated by the transpositions $(12),(34),(56)$, which is isomorphic to $C_2\times C_2\times C_2$.
I have tried to present the argument as rigorously as possible, in keeping with the thorough nature of your solution to the first part. However, I think it's also important to note that the group can be seen in a much simpler, more intuitive way. Basically, $\sqrt{2},\sqrt{3},\sqrt{5}$ are generators of $E$, and they are in some sense independent. Also each one must be mapped by an element of $G$ to either itself, or minus itself (its additive inverse). As such, pretty much all an element of the Galois group does is "flips" some of these square roots. We have three objects to flip (think three levers that can be toggled between two states), so the group is $C_2\times C_2 \times C_2$.
Best Answer
The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$. You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.