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.