Answer amended due to useful comment of Peter Mueller ( so should be unaccepted, as there may be a better one). In a 1994 Journal of Algebra paper, John Thompson and I determined the extension of $\mathbb{Q}$ generated by the values of its irreducible characters. For large enough $n$, this can contain an arbitrarily large number of complex quadratic extensions of $\mathbb{Q}$. However, as Peter Mueller points out, this does not, of itself, answer the question posed here. A key Lemma used in the above paper was a Lemma of James and Kerber ( in Encyclopedia of Mathematics) determines the extension $\mathbb{Q}[\{\chi(g): \chi \in {\rm Irr}(G) \}].$ This is always (at most) a quadratic extension of $\mathbb{Q}$. However, it would take a detailed examination of the proof of that Lemma of James and Kerber to see whether there can be more that two irreducible characters $\chi$ with $\chi(g)$ non-real.
Later edit: Jeremy Rickard explains in his answer exactly what the Lemma of James and Kerber says about the question.