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 $\mathbb{Q}(\sqrt{2},i,\sqrt[8]{3})=\mathbb{Q}(\zeta,\sqrt[8]{3})$.
You've got something wrong: the roots of $t^2+t+1$ are the complex cubic roots of one, not of $-1$: $t^3-1 = (t-1)(t^2+t+1)$, so every root of $t^2+t+1$ satisfies $\alpha^3=1$). That means that you actually want the cubic roots of some of the cubic roots of $1$; that is, you want some ninth roots of $1$ (not of $-1$).
Note that
$$(x^6+x^3+1)(x-1)(x^2+x+1) = x^9-1.$$
So the roots of $x^6+x^3+1$ are all ninth roots of $1$. Moreover, those ninth roots should not be equal to $1$, nor be cubic roots of $1$ (the roots of $x^2+x+1$ are the nonreal cubic roots of $1$): since $x^9-1$ is relatively prime to $(x^9-1)' = 9x^8$, the polynomial $x^9-1$ has no repeated roots. So any root of $x^9-1$ is either a root of $x^6+x^3+1$, or a root of $x^2+x+1$, or a root of $x-1$, but it cannot be a root of two of them.
If $\zeta$ is a primitive ninth root of $1$ (e.g., $\zeta = e^{i2\pi/9}$), then $\zeta^k$ is also a ninth root of $1$ for all $k$; it is a cubic root of $1$ if and only if $3|k$, and it is equal to $1$ if and only if $9|k$. So the roots of $x^6+x^3+1$ are precisely $\zeta$, $\zeta^2$, $\zeta^4$, $\zeta^5$, $\zeta^7$, and $\zeta^8$. They are all contained in $\mathbb{Q}(\zeta)$, which is necessarily contained in the splitting field. Thus, the splitting field is $\mathbb{Q}(\zeta)$, where $\zeta$ is any primitive ninth root of $1$.
Best Answer
The splitting field is either degree $3$ or degree $6$ over $\mathbb{Z}_2$, hence it is either $\mathbb{F}_8$ or $\mathbb{F}_{64}$.
Let $\alpha$ be a root, so that $\mathbb{F}_8 = \mathbb{F}(\alpha)$. The elements are of the form $a+b\alpha+c\alpha^2$, with $\alpha^3=\alpha+1$.
Now, the question is whether any of these elements besides $\alpha$ is a root of the original polynomial $x^3+x+1$.
Note that $(\alpha^2)^3 = (\alpha^3)^2 = (\alpha+1)^2 = \alpha^2+1$, and so if we plug in $\alpha^2$ into the polynomial we have $$\alpha^6 + \alpha^2 + 1 = \alpha^2+1+\alpha^2+1= 0.$$ Thus, $\alpha^2$ is also a root. So the polynomial has at least two roots in $\mathbb{F}_8$, and so splits there.