Degree of a splitting field

It is certainly achievable. For a concrete example:

Let $f(x) \in \mathbb{Q}[x]$ be an irreducible quintic polynomial, and let $K$ be its splitting field. Furthermore, suppose $f$ has exactly $2$ complex roots. It will be convenient to think about $\operatorname{Gal}(K/\mathbb{Q})$ as a permutation group acting on the roots of $f$. Well, $[K:\mathbb{Q}] = | \operatorname{Gal}(K/\mathbb{Q})|$, and since $f$ is irreducible, then $5$ divides $|\operatorname{Gal}(K/\mathbb{Q})|$. Therefore, Cauchy's theorem tells us the Galois group contains an element of order $5$, which is necessarily a $5$-cycle. Next, complex conjugation is also a permutation in the Galois group, and it is a $2$-cycle. It is a theorem that any $2$-cycle together with any $p$-cycle will generate $S_p$ (for $p$ prime). Hence, $\operatorname{Gal}(K/\mathbb{Q}) \cong S_5$, which has order $5!$, and so finally, $[K:\mathbb{Q}]=5!$.

---

For an easier example, try computing the order of the Galois group of any irreducible cubic polynomial in $\mathbb{Q}[x]$ with two complex roots. You'll find that the Galois group is isomorphic to $S_3$ with order $3!$, and hence the degree of the splitting field over $\mathbb{Q}$ is $3!$.

---

Now one might ask whether we can always find extensions of $\mathbb{Q}$ with Galois group $S_n$. Hilbert showed that this is indeed true.

And these are all smaller questions of the more general Inverse Galois Problem: "Does every finite group appear as the Galois group of some Galois extension of $\mathbb{Q}$?", which is not yet known.