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 $\textrm{Inv}(H_3)=\mathbb{Q}(i)$.
Best Answer
I used a "visualization" of this Galois group to find these fixed field. That is, the minimal intermediate fields between $\Bbb{Q}$ and the splitting field $L=\Bbb{Q}(\sqrt{x_1},\sqrt{x_2},\sqrt{x_3})$. Namely, $G=C_2\wr S_3$ is the group of all signed permutations of three coordinates. Or the group of monomial $3\times3$ matrices with a single $\pm1$ on each row/column together with six zeros. Or, the group of symmetries of a cube. To make the linear transfromations related to the monomial matrices symmetries of the cube we can place the vertices of the cube at the points $(\pm1,\pm1,\pm1)$, all the eight sign combinations occur.
I find the last one particularly useful because it allows us to describe the maximal subgroups as permutation of the six roots.
As the OP pointed out we can also view $G$ as a direct product $S_4\times C_2$. As symmetries of the cube the factor $C_2=Z(G)$ is generated by the symmetry $\tau:(x,y,z)\mapsto (-x,-y,-z)$ that obviously commutes with the other linear transformations because its matrix is $-I_3$. Because $\det (-I_3)=-1$ we see that $G$ is the direct product of $C_2=\langle \tau\rangle$ and the subgroup $H$ consisting of the monomial matrices with determinant $+1$ (= the subgroup of orientation preserving symmetries of the cube). The group $G$ acts on the set of four 3-dimensional diagonals of the cube, and it is easy to see that $\tau$ is the only non-trivial symmetry that maps each and every one of those diagonals to itself. Therefore $H\simeq S_4$.
Next we proceed to find the index two subgroups of $G$. They are all normal, so they are the kernels of surjective homomorphisms $f:G\to \{\pm1\}$. To classify such homomorphisms we observe that $G$ is also a Coxeter group generated by the three reflections: $$ \begin{aligned} s_1&:(x,y,z)\mapsto (y,x,z),\\ s_2&:(x,y,z)\mapsto (x,z,y),\\ s_3&:(x,y,z)\mapsto (x,y,-z), \end{aligned} $$ We see that $s_1s_2$ is of order three, so we must choose $f(s_1)=f(s_2)$, but $s_1$ and $s_3$ commute, and $s_2s_3$ has even order, so we can choose $f(s_3)$ independently from $f(s_1)$. The Coxeter relations imply that there are no other constraints to constructing $f$. Therefore we get three different homomorphisms, $f_1,f_2,f_3$, and consequently three different maximal subgroups $H_i=\operatorname{ker}(f_i)$ of index two.
I believe these are all the maximal subgroups of $G$, but I am also prepared to have missed something.
Let's try and identify the corresponding fixed fields.