You are correct in that it is not a splitting field of $\sqrt[4]{2}$, since in particular it is entirely real, but $X^4-2$ is irreducible with complex solutions. Thus it is not a splitting field of any polynomial, since if it were then the extension would be Galois and contain all the roots of $X^4-2$.
One perfectly legitimate strategy would be to look the subfields of some Galois extension $L/\mathbb{Q}$ containing $\mathbb{Q}(\sqrt[4]{2})$ (which can be easily determined once you identify the Galois group) and to identify those that lie inside $\mathbb{Q}(\sqrt[4]{2})$, by looking at the assosciated subgroups of $Gal(L/\mathbb{Q})$ containing $Gal(L/\mathbb{Q}(\sqrt[4]{2}))$. However, this is a lot of work (and requires a lot of theory, although this is not a bad thing in and of itself). There is a much simpler and more elementary way to approach this problem, though.
Hint:
$[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}] = 4$ since $f(X) = X^4-2$ is irreducible, so the only proper subfields are degree $2$ over $\mathbb{Q}$. If $K$ is one such, let $g(X)$ be the minimal polynomial of $\sqrt[4]{2}$ over $K$. $g$ must have degree $2$, divide $f$ and have $\sqrt[4]{2}$ as a root. This gives a small number of possible options. We can then use the fact that coefficients of $g$ lie in $K$ (which is a real field) to determine what possible values $g$ may take, and this tells us enough to be able to determine all possible subfields $K$.
It is worth noting that the actual solution is shorter than the hint, but I think it's a good exercise to work through if you haven't seen it before.
It looks like this problem is leading up to the fundamental theorem of Galois theory, which states that there is a one-to-one correspondence between the subfields of a Galois extension and subgroups of its Galois group. So our answer will be, of course, that no such Galois extension exists (which is why it's difficult to find an example).
First, note that if $K$ is a Galois extension of degree $4$ over $\mathbb{Q}$, then it must be the splitting field of a quartic polynomial. Either that polynomial is irreducible, or it can break into a product of irreducible quadratics. Of course, you'll want convince yourself of these cases and that the cases are indeed exhaustive.
If the polynomial breaks into a product of two irreducible quadratics, then I claim that it's rather easy to show that $K$ must have a subfield.
On the other hand, if the polynomial is irreducible, then $K = \mathbb{Q}[\alpha]$, where $\alpha$ is one of its roots. Since the degree of the Galois extension is $4$, we also know the Galois group has order $4$. As such, the Galois group must have an element $\phi$ of order $2$ that sends $\alpha \mapsto \beta$, where $\beta$ is another root of the polynomial.
Claim: $\mathbb{Q}[\alpha\beta] \subsetneq \mathbb{Q}[\alpha]$
To show this, suppose for contradiction they were actually the same field. Then both would be of degree $4$ over $\mathbb{Q}$, and the former would have a basis $\{1, \alpha\beta, (\alpha\beta)^2, (\alpha\beta)^3\}$. Because $\alpha$ would be an element of both fields, we could write:
$$\alpha = c_1 + c_2(\alpha\beta) + c_3(\alpha\beta)^2 + c_4(\alpha\beta)^3$$
What happens when the automorphism $\phi$ acts on both sides of this expression? What can we conclude?
Best Answer
As you commented $[L:\mathbb Q]=6$. This is because the polynomial $X^3+3$ is irreducible (thanks to Eisenstein criterion, $p=3$) so $\sqrt[3]{-3} = \alpha$ has degree $3$ over $\mathbb Q$.
The splitting field $L$ is $\mathbb Q(\alpha, \omega)$ where $\omega$ is the third primitive root of unity. Since $\omega$ has degree $2$ (it's minimal polynomial is $x^2+x+1$), then $[L:\mathbb Q]=6$ (because $2$ and $3$ are coprime).
Also the extension $L/\mathbb Q$ is normal and separable, hence it is a Galois extension.
Define the following automorphisms of $L$: \begin{equation} \rho:L\rightarrow L, \begin{cases} \rho(\alpha) = \alpha\omega\\ \rho(\omega) = \omega \end{cases} \qquad \sigma:L\rightarrow L, \begin{cases} \sigma(\alpha) = \alpha\\ \sigma(\omega) = \omega^2 \end{cases} \end{equation} They are well defined automorphisms of $\text{Gal}(L/\mathbb Q))$. Now $\rho$ has order $3$, $\sigma$ has order $2$, $\langle \rho\rangle \cap \langle \sigma \rangle = \emptyset$ and $\sigma\rho\sigma^{-1} = \rho^{-1}$. This easily implies (cardinality) that: $$ \text{Gal}(L/\mathbb Q)) = \langle \rho, \sigma \ | \ \rho^3=\sigma^2=id, \sigma\rho\sigma^{-1} = \rho^{-1}\rangle \cong S_3 $$
Thanks to the Galois corrispondance theorem we have that the intermediate subfields are in bijection with the subgroups of $\text{Gal}(L/\mathbb Q)=S_3$.
We have only $6$ subgroups, so we have only $6$ intermediate subfields that are:
\begin{gather} S_3 \quad \longleftrightarrow L^{S_3} = \mathbb Q\\ \langle \rho \rangle \quad \longleftrightarrow L^{\langle \rho \rangle} = \mathbb Q(\omega)\\ \langle \sigma \rangle \quad \longleftrightarrow L^{\langle \sigma \rangle} = \mathbb Q(\alpha)\\ \langle \sigma\rho \rangle \quad \longleftrightarrow L^{\langle \sigma\rho \rangle} = \mathbb Q(\alpha\omega)\\ \langle \sigma\rho^2 \rangle \quad \longleftrightarrow L^{\langle \sigma\rho^2 \rangle} = \mathbb Q(\alpha\omega^2)\\ \{id\} \quad \longleftrightarrow L^{\{id\}} = L\\ \end{gather}
Let's denote with $K$ the splitting field of $X^6+3$ over $\mathbb Q$. $K = \mathbb Q(\sqrt[6]{-3}, \omega_6) = \mathbb (\sqrt[6]{-3},\omega)$ where $\omega_6$ is the sixth primitive root of unity.
Observe that $\left(\sqrt[6]{-3}\right)^2 = \sqrt[3]{-3}$; hence $L\subset K$.
Now if we show that the degree $[K:\mathbb Q]=6$, then we have $L=K$ (a containment and equal degree).
Observe that $\left(\sqrt[6]{-3}\right)^3 = i\sqrt{3}$, and $\mathbb Q(\omega) = \mathbb Q(i\sqrt 3)$. Hence $K=\mathbb Q(\sqrt[6]{-3},\omega) = \mathbb Q(\sqrt[6]{-3})$ and this field has degree $6$ over $\mathbb Q$ because $X^6+3$ is irreducible by Eisenstein.
So $L=K$ and $X^6+3$ splits over $L$.