Let's use the following numbering on the roots of your polynomial: $$\{z_1=\root3\of5,z_2=ωz_1,z_3=ω^2z_1,z_4=\sqrt3,z_5=−\sqrt3\}.$$ Any automorphism $\sigma$ of the splitting field must permute these numbers as they are the zeros of your polynomial. As the splitting field $F$ is generated by them, the automorphism $\sigma$ is fully determined once we know $\sigma(z_j),j=1,2,3,4,5.$ This gives us the usual way of identifying $\sigma$ with an element of $S_5=\operatorname{Sym}(\{z_1,z_2,z_3,z_4,z_5\})$.
But there are further constraints. The numbers $z_1,z_2,z_3$ are zeros of the factor $x^3-5$. As that factor has rational coefficients, all automorphisms must permute these three roots among themselves. Similarly for the remaining pair $z_4,z_5$ as they are the zeros of $x^2-3$. Therefore in the identification of an automorphism with a permutation in $S_5$ of the previous paragraph only the permutations in $\textrm{Sym}(\{z_1,z_2,z_3\})\times \textrm{Sym}(\{z_4,z_5\})$ are allowed. There are 12 such permutations forming a group isomorphic to $S_3\times S_2$. As you had established by other means that $[F:\mathbb{Q}]=12$, we can conclude that all such permutations come from actual automorphisms, and thus the Galois group is isomorphic to $G=S_3\times S_2$.
Let us first calculate the fixed field of the subgroup $H_1=\langle\sigma\rangle$, where $\sigma=(123)(45)$. Here we are given that $\sigma(z_1)=z_2$ and that $\sigma(z_2)=z_3$. As $\sigma$ is an automorphism of fields we get
$$
\sigma(\omega)=\sigma\left(\frac{z_2}{z_1}\right)=\frac{\sigma(z_2)}{\sigma(z_1)}=\frac{\omega^2z_1}{\omega z_1}=\omega.
$$
Therefore $\omega$ is fixed by $\sigma$, and therefore also by any power of $\sigma$.
Thus $\mathbb{Q}(\omega)\subseteq Inv(H_1)$. From Galois theory we know that $[Inv(H_1):\mathbb{Q}]=|G|/|H|=12/6=2$. As $[\mathbb{Q}(\omega):\mathbb{Q}]=2$ it follows that $\mathbb{Q}(\omega)=Inv(H_1)$.
It is fairly clear that the fixed field of the subgroup $H_2\simeq S_3$ that keeps both $z_4$ and $z_5$ fixed is $Inv(H_2)=\mathbb{Q}(\sqrt3)$.
Let's try the last subgroup $H_3$ of order 6.
I describe it as follows. We have a homomorphism $f$ from $S_3=Sym(\{z_1,z_2,z_3\})$ to
$S_2=\textrm{Sym}(\{z_4,z_5\})$ that maps a permutation $\alpha\in S_3$ to the identity element $(4)(5)$ (resp. to the 2-cycle $(45)$) according to whether $\alpha$ is an even (resp. odd) permutation. Then
$$
H_3=\{(\alpha,f(\alpha))\in S_3\times S_2\mid \alpha\in S_3\}.
$$
We easily see that $H_3$ is generated by $\beta=(123)$ and $\gamma=(12)(45)$.
Because $\beta$ acts on the set $\{z_1,z_2,z_3\}$ the same way as $\sigma$ above, we see that $\beta(\omega)=\omega$ and also that $$\beta(\sqrt{-3})=\beta(2\omega+1)=2\omega+1=\sqrt{-3}.$$ Because $\beta(z_4)=z_4$, we obviously also have $\beta(\sqrt3)=\sqrt3$. What about $\gamma$? First we get
$$
\gamma(\omega)=\gamma(\frac{z_2}{z_1})=\frac{\gamma(z_2)}{\gamma(z_1)}=\frac{z_1}{z_2}=\omega^2.
$$
As $2\omega=-1+\sqrt{-3}$ and $2\omega^2=-1-\sqrt{-3}$, this implies that $\gamma(\sqrt{-3})=-\sqrt{-3}$.
But we also have
$$
\gamma(\sqrt3)=\gamma(z_4)=z_5=-\sqrt3.
$$
Putting all these bits together we see that
$$
\gamma(i)=\gamma\left(\frac{\sqrt{-3}}{\sqrt3}\right)=\frac{-\sqrt{-3}}{-\sqrt3}=i.
$$
Similarly we see that $\beta(i)=i$. We can then conclude that $\textrm{Inv}(H_3)=\mathbb{Q}(i)$.
As Ewan Delanoy remarked: pick any $\alpha \in \overline{\mathbb{Q}}$ root of $P=x^4-x-1$.
Then, for $K:=\mathbb{Q}(\alpha)$, you have $[K:\mathbb{Q}]=4$. Now if $[L:K]=2$ and $L/\mathbb{Q}$ is Galois, then $L$ would contain the splitting field of $P=x^4-x-1$.
This is not possible because the splitting field of $P=x^4-x-1$ over ${\mathbb{Q}}$ has degree $4!=24$ which is bigger than $[L:\mathbb{Q}]=8$.
Best Answer
Consider the cyclotomic polynomial $\Phi_7(x) = x^6+x^5+\ldots + x + 1$ which is irreducible and generates an extension of $\Bbb Q$ of degree $6$ which is abelian (i.e. it is Galois with abelian Galois group). Then if $\zeta_7$ is a primitive $7^{th}$ root of $1$, $F=\Bbb Q(\zeta_7)$ is the extension. The element $\zeta_7+\zeta_7^{-1}$ is fixed by complex conjugation (an element of order $2$) and no other automorphism (you can check directly by noting $\zeta_7\mapsto \zeta_7^{k}, 1\le k\le 6$ are the automorphisms of $F$ and that any other automorphism besides $k=6$ gives a different element.
But then $K= \Bbb Q(\zeta_7+\zeta_7^{-1})\subseteq F$ is an extension of degree $3$, because that is the index of the fixing Galois group generated by complex conjugation. Hence $K/\Bbb Q$ is the desired extension. You can even describe it explicitly as $K=\Bbb Q\left((\cos\left({2\pi\over 7}\right)\right)$.
Working out the details you can see it is generated by the polynomial