Let $\Phi=(\xi^{ij})_{0\leq i,j \leq d-1}$. Then $\det\Phi=\prod_{0\leq i<j < d} (\xi^i-\xi^j)=(-1)^{\binom{d}2}\prod_{0\leq i<j < d} (\xi^j-\xi^i)$. So,
\begin{align} \prod_{0\leq i<j < d} (\xi^j-\xi^i)^2
&=\det\begin{bmatrix}1&1 &\dots &1 \\ x_0&x_1&\dots&x_{d-1} \\
\vdots&\vdots&\dots&\vdots \\ x_0^{d-1}&x_1^{d-1}&\dots&x_{d-1}^{d-1} 
\end{bmatrix}
\det\begin{bmatrix}1&x_0 &\dots &x_0^{d-1} \\ 1&x_1&\dots&x_1^{d-1} \\
\vdots&\vdots&\dots&\vdots \\ 1&x_{d-1}&\dots&x_{d-1}^{d-1} \end{bmatrix} \\
&=\det\begin{bmatrix}t_0&t_1 &\dots &t_{d-1} \\ t_1&t_2&\dots&t_d \\
\vdots&\vdots&\dots&\vdots \\ t_{d-1}&t_d&\dots&t_{2(d-1)}\end{bmatrix};
\end{align}
where $x_k=\xi^k$ and $t_k=x_0^k+x_1^k+\cdots+x_{d-1}^k$. Notice that $t_k=0$ unless $k$ is a multiple $d$, in which case it is equal to $d$. Therefore,
$$\prod_{0\leq i<j < d} (\xi^j-\xi^i)^2
=d^d\cdot\det
\begin{bmatrix}1&0&0&\dots&0&0 \\ 0&0&0&\dots&0&1 \\ 
0&0&0&\dots&1&0 \\
\vdots&\vdots&\vdots&\dots&\vdots&\vdots \\
0&1&0&\dots&0&0\end{bmatrix}=d^d(-1)^{\binom{d-1}2}.$$
We gather that
$$\det\Phi=(-1)^{\binom{d}2}\cdot d^{\frac{d}2}\cdot i^{\binom{d-1}2}
=d^{\frac{d}2}\cdot i^{\frac{(3d-2)(d-1)}2}.$$

A couple of noteworthy facts:  if $V=\prod_{0\leq i<j < d} (\xi^j-\xi^i)$ denote the Vandermonde determinant, then 

$V^2$ is the discriminant of the polynomial $f(z)=z^d-1$;

$V^2$ is a symmetric polynomial and when expanded in elementary polynomials basis then only $e_d(\xi_0,\dots,\xi_{d-1})$ survives.

By the way, in general, computing the coefficients of even-powers of the Vandermonde's determinant $V(y_1,\dots,y_n)$ symmetric polynomials basis is a difficult and big sport in mathematics and physics.

**UPDATE.** This is in response to EFinat-S's quest on replacing $\xi$ by an indeterminate $x$. The determinant is just
$$\Delta(d)=\det\left[x^{ij}-1\right]_{i,j}^{1,d}
=\prod_{0\leq i<j\leq d}\left(x^j-x^i\right).$$