Numerical experiment (as far as $d=50$) makes it clear that $\Delta(d)=d^{d/2}i^{m_d}$ where $$ m_d = \begin{cases} 0 & \text{ if } d = 1 \text{ or } 6 \pmod{8} \\ 1 & \text{ if } d = 0 \text{ or } 7 \pmod{8} \\ 2 & \text{ if } d = 2 \text{ or } 5 \pmod{8} \\ 3 & \text{ if } d = 3 \text{ or } 4 \pmod{8}. \end{cases} $$ I haven't tried to find a proof, however.
Neil Strickland
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