Let $\Lambda$ be a full-rank lattice in $\mathbb{R}^n$, with basis $v_1,\ldots, v_n$. The ratio $\frac{\prod_{i=1}^n\|v_i\|}{|\det\Lambda|}$ is sometimes called the Hadamard ratio as Hadamard proved that this ratio is smaller than 1 (with equality when the basis is orthogonal). This ratio can be used as a measure (or defect from) orthogonality of a basis for a lattice.
I want to know about the "worst" lattices, i.e. in a given dimension, what is the value of $$ \sup_{\Lambda\leq\mathbb{R}^n} \ \inf_{\Lambda=\langle v_1,\ldots, v_n\rangle_{\mathbb{Z}}} \ \frac{\prod_{i=1}^n\|v_i\|}{|\det\Lambda|} $$ and what lattices obtain this value?
In dimension two, I believe the hexagonal lattice is the worst at $2/\sqrt{3}$, but other than that I don't know (maybe lattices built from the standard $n$-simplex are bad?).
If not exact answers, perhaps asymptotic behavior, or bounds (as functions of $n$)?
The $n$-simplex comment is conjecturing the sup-inf is given by $$ \frac{2^{n/2}}{\sqrt{n}}=1, \ 2/\sqrt{3}, \ \sqrt{2}, \ 2, \ \ldots $$ The rows of the matrix below showing the iterative construction going from dimension $3$ to $4$ $$ \left( \begin{array}{ccc|c} 1&0&0&0\\ \frac{1}{2}&\frac{\sqrt{3}}{2}&0&0\\ \frac{1}{2}&\frac{1}{2\sqrt{3}}&\sqrt{\frac{2}{3}}&0\\ \hline a&b&c&\sqrt{1-(a^2+b^2+c^2)} \end{array} \right) $$ where $(a,b,c)$ is the centroid of $(0,0,0)$ and the three rows above it, and the last entry is to make the row unit length.
The guess the lattice I'm referring to is the $A_n$ root lattice. Is this the least orthonormal lattice in some sense?