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 $$\Delta(d)=\det\left[x^{ij}-1\right]_{i,j}^{1,d} =\prod_{0\leq i<j\leq d}\left(x^j-x^i\right).$$

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 $$\Delta(d)=\det\left[x^{ij}-1\right]_{i,j}^{1,d} =\prod_{0\leq i<j\leq d}\left(x^j-x^i\right).$$

POSTSCRIPT. We can generalize the determinant evaluation as follows. Start with the Vandermonde $$\prod_{i<j}^{\pmb{0,d-1}}(y_j-y_i)=\det\begin{bmatrix}1&y_0 &\dots &y_0^{d-1} \\ 1&y_1&\dots&y_1^{d-1} \\ \vdots&\vdots&\dots&\vdots \\ 1&y_{d-1}&\dots&y_{d-1}^{d-1} \end{bmatrix},$$ then choose $y_0=1$ and subtract the first column from the rest to get $$\det\begin{bmatrix}1&y_0-1 &\dots &y_0^{d-1}-1 \\ 1&y_1-1&\dots&y_1^{d-1}-1 \\ \vdots&\vdots&\dots&\vdots \\ 1&y_{d-1}-1&\dots&y_{d-1}^{d-1}-1 \end{bmatrix} =\det\begin{bmatrix}1&y_1-1 &\dots &y_1^{d-1}-1 \\ 1&y_2-1&\dots&y_2^{d-1}-1 \\ \vdots&\vdots&\dots&\vdots \\ 1&y_{d-1}-1&\dots&y_{d-1}^{d-1}-1 \end{bmatrix}.$$ Consequently, $$\det\begin{bmatrix}1&y_1-1 &\dots &y_1^{d-1}-1 \\ 1&y_2-1&\dots&y_2^{d-1}-1 \\ \vdots&\vdots&\dots&\vdots \\ 1&y_{d-1}-1&\dots&y_{d-1}^{d-1}-1 \end{bmatrix} =\prod_{k=1}^{d-1}(y_k-1)\cdot\prod_{i<j}^{\pmb{1,d-1}}(y_j-y_i).$$ The latter determinant is in the form of your question, with $y_j=\xi^j$.

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.

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 $$\Delta(d)=\det\left[x^{ij}-1\right]_{i,j}^{1,d} =\prod_{0\leq i<j\leq d}\left(x^j-x^i\right).$$