Let $A$ be the character table as a square matrix. In other words $A_{ij}=\chi_i(g_j)$, where $\chi_i$ the distinct irreducible characters, and $g_j$ are representatives of conjugacy classes. Let $A^H$ be the conjugate transpose. Let us compute the matrix product
$B:=A^HA$. At position $(i,j)$ we get
$$
B_{ij}=\sum_k A^H_{ik}A_{kj}=\sum_k \overline{A_{ki}}A_{kj}=\sum_k\overline{\chi_k(g_i)}\chi_k(g_j)=|C_G(g_j)|\delta_{ij}
$$
by the second orthogonality relation. Therefore $B$ is diagonal, and
$$
\det B=\prod_j|C_G(g_j)|.
$$
So by the multiplicativity of determinant we also have
$$
\det(A^H)\det(A)=\det B=\prod_j|C_G(g_j)|.
$$
Let us study the relation of $A^H$ and the transpose $A^T$.
Let us define a permutation $s$ of the conjugacy classes by the mapping $[g]\mapsto [g^{-1}]$.
It is obvious a product of disjoint 2-cycles, and its fixed points are exactly the conjugcay classes stable under taking the inverse element. Let us denote by $\ell$
the number of orbits of size two.
If we denote by $\tilde{A}$ the matrix that we get from $A$ by permuting the columns according to $s$, then the general fact $\chi(g^{-1})=\overline{\chi(g)}$ allows us to identify $A^H$ as $\tilde{A}^T$. Clearly
$$
\det\tilde{A}=(-1)^{\ell}\det A,$$
so we get the equation
$$
(-1)^{\ell}(\det A)^2=\det B=\prod_j|C_G(g_j)|.
$$
The sign of $\det A$ will always remain ambiguous, because we have no natural ordering neither for the conjugacy classes nor for the characters, so all we can say is that
$$
\det A=\pm i^{\ell}\sqrt{\prod_j|C_G(g_j)|}.
$$
