Using the earlier responses and comments, I confirm the formula suggested by Neil Strickland:
$$\Delta(d)=d^{d/2}i^{m(d)}\qquad\text{with}\qquad m(d): = 1 + d(7-d)/2\in\mathbb{Z}.$$
Consider the $d\times d$ Vandermonde matrix
$$\Phi(d):=(\xi^{ij})_{0\leq i,j \leq d-1}.$$
Subtracting the first column from each other column, we get a matrix with first row equal to $(1,0,\dots,0)$ and lower $(d-1)\times(d-1)$ block equal to the OP's matrix. Therefore,
$$\Delta(d)=\det\Phi(d).$$
It is straightforward to check that $\Phi(d)^\ast\cdot\Phi(d)$ equals $d$ times the identity matrix, therefore
$$ |\det\Phi(d)|^2=d^d.$$
In other words, $|\det\Phi(d)|=d^{d/2}$, and we are left with determining
$$\frac{\det\Phi(d)}{|\det\Phi(d)|}=\prod_{0\leq i<j\leq d-1}\frac{\xi^j-\xi^i}{|\xi^j-\xi^i|}.$$
Let me use the notation $e(t):=e^{2\pi it}$, familiar from analytic number theory. Then we see that
$$\xi^j-\xi^i=e\left(\frac{j}{d}\right)-e\left(\frac{i}{d}\right)
=e\left(\frac{i+j}{2d}\right)\left(e\left(\frac{j-i}{2d}\right)-e\left(\frac{i-j}{2d}\right)\right).$$
On the right hand side, $0<\frac{j-i}{2d}<\frac{1}{2}$, hence $e\left(\frac{j-i}{2d}\right)$ lies in the upper half-plane. As a result,
$$\frac{\xi^j-\xi^i}{|\xi^j-\xi^i|}=e\left(\frac{i+j}{2d}\right)i.$$
We need to calculate the product of the right hand side over the $\binom{d}{2}$ pairs $0\leq i<j\leq d-1$. By symmetry (or by brute-force calculation), the average of $i+j$ equals $d-1$, whence
$$\prod_{0\leq i<j\leq d-1}\frac{\xi^j-\xi^i}{|\xi^j-\xi^i|}=\left(e\left(\frac{d-1}{2d}\right)i\right)^{\binom{d}{2}}=e\left(\left(\frac{d-1}{2d}+\frac{1}{4}\right)\binom{d}{2}\right).$$
We calculate
$$\left(\frac{d-1}{2d}+\frac{1}{4}\right)\binom{d}{2}=\frac{(3d-2)(d-1)}{8},$$
therefore in the end
$$\Delta(d)=d^{d/2}i^{n(d)}\qquad\text{with}\qquad n(d):=(3d-2)(d-1)/2\in\mathbb{Z}.$$
This agrees with Neil Strickland's formula, upon noting that $m(d)\equiv n(d)\pmod{4}$, i.e.,
$$2+d(7-d)\equiv (3d-2)(d-1)\pmod{8}.$$

**Added.** As Alexey Ustinov remarked, $\Phi(d)$ is known as a Schur matrix. As Carlitz wrote [in his 1959 Acta Arithmetica paper][1], "this matrix is familiar in connection with Schur's derivation of the value of Gauss's sum". In fact, on page 295 of this paper, Carlitz uses the known value of Gauss's sum to find the eigenvalues of this matrix (which are all of the form $\pm\sqrt{d}$ and $\pm i\sqrt{d}$, hence one only needs to find the 4 multiplicities). This can be regarded as a refinement and an alternate proof of the above result, since the product of the eigenvalues is the determinant. 


  [1]: http://matwbn.icm.edu.pl/ksiazki/aa/aa5/aa536.pdf