No I don't think this is right. For example the extension $\Bbb{Q}(\sqrt{3},\sqrt{2})$ is a Galois extension of $\Bbb{Q}$ because it is the splitting field of $(x^2 - 3)(x^2-2)$. This polynomial has exactly 4 distinct roots in $\Bbb{Q}(\sqrt{3},\sqrt{2})$ and it can easily be shown that its Galois group is the Klein 4-group $V_4$. But $V_4$ has order 4 that is not equal to $4!=24$.
Now you are thinking that the Galois group has to be $S_4$. Let me tell you why this is not possible. Let us call $\sqrt{3}$ root #1, $\sqrt{2}$ root #2, $-\sqrt{3}$ root #3, $-\sqrt{2}$ root #4. Given a cycle in $S_4$ let that cycle act on the roots simply by permuting the numbers $1,2,3$ and $4$. For example, the cycle $(12)$ exchanges $\sqrt{3}$ and $\sqrt{2}$ and keeps the negative guys fixed. But then this cannot possibly be a valid element of the Galois group because:
Elements of the Galois group must send for example $\sqrt{2}$ to another root of the minimal polynomial of $\sqrt{2}$ over $\Bbb{Q}$. The minimal polynomial of $\sqrt{2}$ when viewed as an element of $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is $x^2 -2$. Therefore the only possibility for where $\sqrt{2}$ can be sent to is $-\sqrt{2}$, because this is the only other root of this polynomial in any splitting field. So therefore the cycle $(12)$ above cannot be a valid element of the Galois group (if we view the Galois group as sitting inside of $S_4$).
It follows that the Galois group in this case can only be viewed as a proper subgroup of $S_4$ and hence cannot have order $4!=24$.
Edit: Since you seem to be having some trouble understanding the Galois group, let me explain a bit more here. Now I assume that you know what an $F$ - algebra is (otherwise how would you understand field extensions?)
The following is the start of how one describes the Galois group:
Let $A = F[x]$ where $F$ is a field. Let $\iota_A$ denote inclusion of $F$ into $A$. Then any $F$ - algebra homomorphism from $F[x]$ to some other $F$ - algebra $B$ (where the homomorphism in question for $B$ is just the inclusion map $\iota_B : F \to B$) is completely determined by specifying the image of $x$ in $B$. This is because if we have an $F$ - algebra homomorphism $\varphi$ from $F[x] \rightarrow B$, we must have that
$$\iota_B = \varphi \circ \iota_A.$$
In particular this means that $\varphi$ must be the identity on the coefficients of a polynomial in $F[x]$. This explains why $\varphi$ is completely determined by its action on $x$. Now we claim that we have a bijection of sets
$$\Big\{\operatorname{Hom}_{\text{$F$ -algebra}} \big(F[x],B\big)\Big\} \longleftrightarrow B $$
where the bijection is given by $f$ that maps $\varphi$ on the left to $\varphi(x)$ with inverse $g$ that maps an element in $b \in B$ to the homomorphism $\varphi_b$ which is evaluation at $b$. Viz. $\varphi_b$ is just the homomorphism that sends $x$ to $b$. You can check that $f$ and $g$ are mutual inverses.
Now a corollary of this is that we have a bijection
$$\Big\{\operatorname{Hom}_{\text{$F$ -algebra}} \big(F[x]/(f(x)),B\big)\Big\} \longleftrightarrow \Big\{b\in B : \varphi_b(f(x)) = 0 \Big\}. $$
I will leave you to work out the details of how this comes from the fact I stated before. Essentially it is due to the universal property of quotient rings that says given a unique ring homomorphism $\varphi$ from $F[x]$ to $B$ we have unique ring homomorphism from the quotient of $F[x]/(\ker \varphi)$to $B$ such that $\varphi$ factorises through the quotient. I can edit my post to explain this more if you wish.
This is the start of how one gets a description of the Galois group because giving a homomorphism from some field say $F(\alpha)$ to itself (which is automatically an automorphism by the Rank - Nullity Theorem) is by my description above equivalent equivalent to specifying a root of the minimal polynomial of $\alpha$ over $F$ in $B$. But then our $B$ here is exactly what we started with, that is $F(\alpha)$ so that $\alpha$ must be sent to another root of its minimal polynomial over $F$.
Does this help to explain more to you?
Note that any automorphism $\sigma$ must send $\alpha\in \mathbb Q(\sqrt2,\sqrt3,\sqrt5)$ to another root of the minimal polynomial of $\alpha$. In particular, $\sigma\sqrt2=\pm \sqrt2,\sigma\sqrt3=\pm\sqrt3,$ and $\sigma\sqrt5=\pm\sqrt5$. Since $\{\sqrt2,\sqrt3,\sqrt5\}$ is a generating set for $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$ as a field over $\mathbb Q$, we see that $\sigma$ is uniquely determined by $\sigma\sqrt2,\sigma\sqrt3,$ and $\sigma\sqrt5$. Thus the only possible automorphisms are
$$\begin{align}
a+b\sqrt2+c\sqrt3+d\sqrt5&\mapsto a+b\sqrt2+c\sqrt3+d\sqrt5\\
a+b\sqrt2+c\sqrt3+d\sqrt5&\mapsto a-b\sqrt2+c\sqrt3+d\sqrt5\\
&\vdots\\
a+b\sqrt2+c\sqrt3+d\sqrt5&\mapsto a-b\sqrt2-c\sqrt3-d\sqrt5\\
\end{align}$$
and in order to show that these are all automorphisms, it suffices to show that there are exactly $8$ automorphisms of $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$. Since $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$ is the splitting field of a polynomial, we know that it is Galois, so the number of automorphisms it has is equal to its degree. Recall that
$$[\mathbb Q(\sqrt2,\sqrt3,\sqrt5):\mathbb Q]=[\mathbb Q(\sqrt2,\sqrt3,\sqrt5):\mathbb Q(\sqrt2,\sqrt3)][\mathbb Q(\sqrt2,\sqrt3):\mathbb Q(\sqrt2)][\mathbb Q(\sqrt2):\mathbb Q]$$
and so it suffices to show that each degree on the RHS is $2$. Clearly they are at most $2$, as each extension is obtained by adjoining a root of a quadratic polynomial. Thus it suffices to show that each extension on the RHS is nontrivial, i.e. that $\sqrt2\notin \mathbb Q,\sqrt3\notin \mathbb Q(\sqrt2)$ and $\sqrt5\notin \mathbb Q(\sqrt2,\sqrt3)$. The first is a famous theorem. Since $\mathbb Q(\sqrt2)$ has basis $\{1,\sqrt2\}$, if $\sqrt3\in \mathbb Q(\sqrt2)$ we would have $\sqrt3=a+b\sqrt2$ with $a,b\in\mathbb Q$, so $3=a^2+2b^2+2ab\sqrt2$. Since $\sqrt2$ is irrational we must have $2ab=0$, so $a=0$ or $b=0$, thus we need only observe that $3$ and $3/2$ are not squares in $\mathbb Q$. The same technique (with some additional effort) works to show that $\sqrt5\notin \mathbb Q (\sqrt2,\sqrt3)$, observing that $\{1,\sqrt2,\sqrt3,\sqrt6\}$ is a basis for $\mathbb Q(\sqrt2,\sqrt3)$.
Once you see that these are the automorphisms, it should be relatively easy to see what their fixed fields are. For example, the map
$$a+b\sqrt2+c\sqrt3+d\sqrt5\mapsto a-b\sqrt2+c\sqrt3+d\sqrt5$$
has fixed field $\mathbb Q(\sqrt3,\sqrt5)$ while the map
$$a+b\sqrt2+c\sqrt3+d\sqrt5\mapsto a-b\sqrt2-c\sqrt3+d\sqrt5$$
has fixed field $\mathbb Q(\sqrt6,\sqrt5)$ (why $\sqrt6$?). These fixed fields are all the maximal subfields of $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$, and the remaining intersections are pairwise intersections of these subfields (since the only other nontrivial subgroups of the Galois group are generated by pairs of automorphisms), which are easy to determine.
Best Answer
Since $x^4-2$ is irreducible over $\mathbb{Q}$ (Eisensein's Criterion at $p-2$, for example), the Galois group is transitive on the four roots. There is an automorphism that maps $\sqrt[4]{2}$ to $i\sqrt[4]{2}$; call it $\rho$. This map either maps $i$ to $i$, or $i$ to $-i$.
If it maps $i$ to $i$, then consider $\sigma$, complex conjugation. We have that $\sigma\rho$ maps $i$ to $-i$ and $\sqrt[4]{2}$ to $-i\sqrt[4]{2}$, whereas $\rho\sigma$ maps $i$ to $-i$ but $\sqrt[4]{2}$ to $i\sqrt[4]{2}$. So $G$ is not abelian.
If $\rho$ maps $i$ to $-i$, then $\sigma\rho$ maps $\sqrt[4]{2}$ to $-i\sqrt[4]{2}$ and $i$ to $i$. Taking $(\sigma\rho)^3$ we obtain a map that sends $i$ to $i$ and $\sqrt[4]{2}$ to $i\sqrt[4]{2}$, so we are back in the previous case. So either way, $G$ is not abelian.
That means that it is either dihedral, or the quaternion group of order $8$ (the only nonabelian groups of order $8$). But in the quaternion group, the only element of order $2$ is central (it is $-1$), whereas complex conjugation, which is of order $2$ in the Galois group, is not central, as we just saw. That means that the Galois group must be the dihedral group of order $8$. Alternatively, in the quaternion group there is a single element of order $2$; but $G$ has at least two: complex conjugation and $\rho^2$. So $G$ must be dihedral, not quaternion.
Explicitly, our map $\rho$ that sends $\sqrt[4]{2}$ to $i\sqrt[4]{2}$ and $i$ to $i$ is of order $4$; complex conjugation $\sigma$ is of order $2$; now note that $\sigma\rho=\rho^3\sigma$ (both map $i$ to $-i$, and $\sqrt[4]{2}$ to $-i\sqrt[4]{2}$). This gives you explicitly the dihedral structure: $G$ contains $\langle \sigma,\rho\mid \sigma^2=\rho^4=1,\ \sigma\rho=\rho^3\sigma\rangle$, which is of order $8$, so $G$ is this group, which is the dihedral group of order $8$.