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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 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.

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.

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 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. If negative, this answers the question (about the columns). If positive, the corresponding question about the rows would need to be analysed. Unfortunately, I can't access this article of James and Kerber at the moment but hopefully others can.

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 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.

No, this is certainly not the case in general: In the paper "Sums of squares and the fields $\mathbb{Q}_{A_{n}}$" ( Journal of Algebra, 1994), John Thompson and I proved that for $n > 24$, the field generated by the complex character values of $A_{n}$ is $\mathbb{Q}[\{ \sqrt{\epsilon(p)p} \}]$, where $p$ runs through odd primes $p < n$ with $p \neq n-2$, and $\epsilon(p)$ is the sign with $p \equiv \epsilon(p)$ (mod $4$).

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 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. If negative, this answers the question (about the columns). If positive, the corresponding question about the rows would need to be analysed. Unfortunately, I can't access this article of James and Kerber at the moment but hopefully others can.