In $\Bbb F_5[X]$
What we know
We have that $f = X^6 - 1 = (X-1)(X+1)(X^4 + X^2 +1)$, so $Ω^f_{\Bbb F_5} = Ω^{X^4 + X^2 + 1}_{\Bbb F_5}$. Evaluating $f$ in all elements of $\Bbb F_5$ shows there are no other roots of $f$ there. I don't believe that allows us to conclude that $X^4 + X^2 +1$ is therefore irreducible, though (although I'm not exactly sure why not). Moving on, we can invoke the quadratic formula and find, with some diligence, that the roots of the quartic are $\pm\sqrt{2 \pm 3\sqrt{2}}$. This is a little messy, though, so I don't think this is the right course of action.
Where we're going
The Galois group of a finite extension $L/K$ of a finite field $K$ is cyclic of order $[L:K]$, generated by $\left(x ↦ x^{\#K}\right)$. So all we need to know is $\left[\Omega^{X^4 + X^2 + 1}_{\Bbb F_5} : \Bbb F_5\right]$. The obvious guess would of course be 4, but as I'm not sure if the quartic is irreducible, or if the answer even depends on its irreducibility, I cannot say for certain. One more thing I know about finite fields is that if we have one root, say $\alpha = \sqrt{2 + 3\sqrt{2}}$, then the other roots are just powers of it and are therefore in $\Bbb F_5(\alpha)\cong \Bbb F_5[X]/(f^{\alpha}_{\Bbb F_5 })$. This shows that the sought-after degree equals deg$f^{\alpha}_{\Bbb F_5}$ which makes me realise that if the quartic is indeed irreducible - and therefore $\alpha$'s minimal polynomial - we would indeed be done (Gal$(f) ≅ C_4$).
Can anyone help me try to make these ends meet?
In $\Bbb F_7[X]$
Here $f = X^6 + 1$ has no roots in the base field. Again the (or at least: a) problem lies in determining irreducibility.
in $F_5[x]$ it is $X^6-1$ whose splitting field is $F_{5^n}$ where $n$ is the least integer such that $6 |5^n-1$. The Galois group is generated by the Frobenius which has order $n$.
In $F_7[x]$ it is $X^6+1$ whose splitting field is the same as $X^{12}-1$ ie. $F_{7^m}$ where $m$ is the least integer such that $12 | 7^m-1$.