determinant of the table of characters I am certain that the answer to this question exists somewhere. It might be a classical exercise.
Let $G$ be a finite group. Its table of characters is a square matrix, whose rows are indexed by the conjugacy classes and the columns are indexed by the irreducible characters. It is well defined, up to the order of rows and columns. In particular, its determinant if well-defined up to the sign. Let us define $\Delta$ to be the square of this determinant (this is well-defined). Because the characters form a basis of the space of class functions, we know that $\Delta\ne0$. When $G={\mathbb Z}/n{\mathbb Z}$, $\Delta=n^n$.

Is there a close formula for $\Delta$ for a general group? Is it always an integer?

 A: 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.
A: Hasn't this been discussed at length in the recent Annals paper
An Elementary Exposition of Frobenius's Theory of Group-Characters and Group-Determinants
Leonard Eugene Dickson
(1902)?
The Determinant the OP discusses is NOT the group determinant per se, but comes up shortly after the definition of the group determinant.
A: Corrected version
I believe it is always an integer.  We can assume that all our representations are over the algebraic closure $\overline{\mathbb Q}$ of $\mathbb Q$.  If $\Gamma$ is the absolute Galois group, then clearly $\Gamma$ acts on the characters of $G$ (if you have a representation, then twist it by the action of $\Gamma$).  It thus follows that the determinant squared is fixed by $\Gamma$ (since $\Gamma$ permutes the rows) and so is a rational number.  But it is also an algebraic integer so it is an integer.  
A: If $A$ is the character table and $A^\ast$ is its conjugate transpose, then the orthogonality relations tell us that $A A^\ast = \text{diag}\{|C_G(g)|\} $, where the enties run over a fixed choice of elements of $G$, one from each conjugacy class. Thus $|\Delta| = \det A A^\ast = \prod |C_G(g)|$ is an integer. On the other hand, $\Delta$ must be rational. This follows from the fact that the action of $\text{Gal}(\overline{\mathbb Q}/\mathbb Q)$ permutes the columns of $A$, hence fixes $\Delta  = (\det A)^2$. Thus $\Delta=\pm |\Delta|$ is an integer.
