I'm trying to find the minimal polynomial of $\epsilon$ over $\mathbb{Q}(\epsilon^3)$ so the degree of the extension is equal to the degree of the minimal polynomial.
I am thinking maybe $\epsilon^3x^3-1=0$ to be the minimal polynomial but it is not monic so it would be $x^3-1/\epsilon^3=0$. Does this make sense?
It would work, but it's not the most natural choice.
If you have rational numbers as well as $\epsilon^3$ at your disposal, what's the simplest polynomial you can make that has $\epsilon$ as a root? It's $x^3-\epsilon^3$. So that's what you're after.
It's worth noting that $\frac1{\epsilon^3} = \bar{\epsilon}^3 = \epsilon^6$ is the other primitive cube root of unity. So your polynomial would be the minimal polynomial of the other three primitive ninth roots of unity that aren't roots of $x^3 - \epsilon^3$. The splitting field of the two polynomials, and thus the related extension degrees, are the same, as any field that contains one primitive ninth root of unity contains all of them.