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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., $$2+d(7-d)\equiv (3d-2)(d-1)\pmod{8}$$