That is because there is a theorem that states that in every perfect field ($\mathbb Q$ and $\mathbb F_p$ are examples), every irreducible polynomial is separable (i.e. all their roots have multiplicity 1). Your polynomial f(x) in your assertion is not necessarily separable, but if you write $ f(x) = (p_1(x))^{a_1} (p_2(x))^{a_2}... (p_n(x))^{a_n}, \quad a_i \ge 1 $ with $p_i(x)$ being the irreducible factors of $f$, then F is also the splitting field for $ g(x) = p_1(x) p_2(x) ... p_n(x). $ since both of these polynomials ($f$ and $g$) have exactly the same roots. The polynomial g is separable since it is a product of irreducible polynomials (if you use the Dummit & Foote, which you should if you don't, I recommend ; I am speaking of Corollary 34 page 547.) so that every root of $g$ has multiplicity 1.
Therefore, the splitting field of $f$ being Galois over $\mathbb Q$ doesn't mean that $f$ is separable, but rather that some polynomial (the $g$ one) divides $f$ and has the same roots as $f$, with $g$ being separable and having the same splitting field as $f$.
I hope this helped.