Character Values for Alternating Groups of degree $\geq 7$
As Geoff thought, the answer is contained in James and Kerber (it's Theorem 2.5.13 in "The Representation Theory of the Symmetric Group", Encyclopedia of Mathematics and its Applications vol. 16, 1981).
Each row or column of the character table contains at most one pair of irrationalities.
If $\lambda$ is a self-conjugate partition (and so corresponds to an irreducible character of $S_n$ that splits on restriction to $A_n$), then the only cycle type of elements where the corresponding character values are irrational is the one with cycle lengths equal to the lengths of the hooks centred on the main diagonal of the Young diagram (notice that these are all odd and distinct, so correspond to a conjugacy class of $S_n$ that splits in $A_n$).
For example, for the partition $(6,6,6,6,6,6)$ this cycle type would be $[11,9,7,5,3,1]$.
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.
As a related question, one can ask how many of the irreducible characters have only real values. It is known that this is the same as the number of real conjugacy classes, where $[x]$ is said to be real iff it is the same as $[x^{-1}]$. Let $G$ be the alternating group on $n$ letters, let $S$ denote the corresponding symmetric group, and let $\epsilon\colon S\to\mathbb{Z}/2$ be the map with kernel $G$. If $x\in G$ then there is an obvious way (starting from a disjoint cycle decomposition) to construct $y\in S$ with $yxy^{-1}=x^{-1}$. If $m_i$ is the number of $i$-cycles in $x$ then it works out that $$ \epsilon(y) = \sum_im_i\lfloor i/2\rfloor = \sum_j (m_{4j+2}+m_{4j+3}). $$ If $\epsilon(y)=0$ then clearly $[x]$ is real. If $\epsilon(y)=1$ then $[x]$ is real iff there is an odd permutation $z$ that commutes with $x$ (so we can replace $y$ by $yz$). I think that this holds iff there is an even $i$ with $m_i>0$, or an odd $i$ with $m_i>1$. Thus, we see that $[x]$ is non-real iff it is a product of cycles of distinct odd lengths, an odd number of which are congruent to $3$ mod $4$. There will be many classes like this when $n$ is large, so there will be many non-real irreducible characters. It seems unlikely to me that each such character has only two non-real values, but I do not see a proof. Anyway, I hope that this analysis may at least shed some light on the original question.