Denote by $K$ the splitting field of $x^8-1$. Clearly $x^8-1$ has $8$ complex roots, namely $\zeta^i\cdot\sqrt[8]{3}$ for $i=0,\dots,7$ and $\zeta$ is a primitive $8^{th}$ root of unity. Since $K$ contains $\zeta\cdot\sqrt[8]{3}$ and $\sqrt[8]{3}$, it must contain their quotient $\zeta$. It also contains $\sqrt[8]{3}$, and we can conclude $K\supseteq\mathbb{Q}(\zeta,\sqrt[8]{3})$. On the other hand, any field containing these two elements must contain the splitting field for $x^8-1$, hence $K=\mathbb{Q}(\zeta,\sqrt[8]{3})$.
To determine the dimension of this extension, recall that field automorphisms permute the roots of minimal polynomials. $\sqrt[8]{3}$ has minimal polynomial $x^8-3$ over $\mathbb{Q}$, ($x^8-3$ is 3-Eisenstein, hence irreducible). $\zeta$ has minimal polynomial $\Phi_8(x)=x^4+1$ (see below), hence $\mathbb{Q}(\zeta)$ is an extension of dimension 4 over $\mathbb{Q}$. It is not difficult to see that $\zeta\notin\mathbb{Q}(\sqrt[8]{3})$, hence $K$ is of dimension $4\cdot8=32$ over $\mathbb{Q}$.
Last, recall (see Dummit & Foote 13.6, for instance) that the $n^{th}$ roots of unity satisfy
$x^n-1=\prod\Phi_d(x)$
where $d$ goes over all the divisors of $n$. In our case, $n=8$,
$x^8-1 = (x^4-1)\cdot(x^4+1)=(x-1)(x+1)(x^2+1)\cdot(x^4+1)$
where the first three multipliers at the last equality above are $\Phi_1(x)=x-1$, $\Phi_2(x)=x+1$ and $\Phi_4(x)=x^2+1$.
As a final remark, we could have taken $\zeta=(1+i)\sqrt{2}/2$. From a similar argument to before, $\mathbb{Q}(\sqrt{2},i)\supseteq\mathbb{Q}(\zeta)$. However, both sides are a 4-dimensional extension over $\mathbb{Q}$, hence they must be equal. This can also be proven directly, by writing $i$ and $\sqrt{2}$ only using powers of $\zeta$.
To summarize, the splitting field of $x^8-3$ is $\mathbb{Q}(\sqrt{2},i,\sqrt[8]{3})=\mathbb{Q}(\zeta,\sqrt[8]{3})$.