I found the following answer after posting it:
$$\Delta=\epsilon\prod_c\frac{|G|}{|c|},\qquad\epsilon=(-1)^m,$$
where the product is taken over the conjugacy class. And $m$ is the number of pairs of complex conjugate irreducible characters.

*Proof*. On the one hand, the complex conjugate of the table is itself, up to $m$ transpositions of rows. This is because the conjugate of an IC is an IC. Therefore
$$\overline{\det(TC)}=\epsilon\det(TC)$$
($TC$ stands for ``table of characters''.)
Hence $\det(TC)$ is real if $m$ is even, pure imaginary if $m$ is odd. hence $\Delta$ is real and its sign is $\epsilon$.

Now the characters form a unitary basis. Because a unitary matrix has a unit determinant, we may compute $|\Delta|$ by taking any unitary basis. Take $\phi_c(g)$ to be $0$ if $g\not\in c$ and $|G|^{1/2}/|c|^{1/2}$ if $g\in c$. In particular $|\Delta|$ is an integer because $$\frac{|G|}{|c|}=|{\mathcal Z}(a)|,\qquad a\in c.$$

*Another Proof*: Let $D$ be the diagonal matrix whose diagonal entries are the cardinals of the congugacy classes. We may assume that the first rows of $TC$ are the real characters and the $2m$ last ones are the pairs of complex conjugate characters. Then the $(i,j)$-entry of $M:=(TC)D(TC)^T$ is $|G|\langle\overline{\chi_i},\chi_j\rangle$. From the orthogonality relations, we see that $M={\rm diag}(1,\ldots,1,J,\ldots,J)$ where
$$J=\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}.$$
The number of blocks $J$ is precisely $m$. Now take the determinant; we obtain $\Delta\det D=(-1)^m|G|^r$ where $r\times r$ is the size of $TC$. Hence the formula.

submatricesof the character table, a much more subtle question. See front.math.ucdavis.edu/1110.0818. – Richard Stanley Oct 14 '11 at 0:22