Suppose $\Lambda\subset\mathbb{R}^n$ is a unimodular (i.e. volume $1$) lattice in Euclidean space. Let $v_1,\dots, v_n\in\Lambda$ be a basis of $\Lambda$ such that the product of lengths $A=v_1\cdotsv_n$ is minimal. I'd like to bound this minimal product $A$ from above as $\Lambda$ varies (and $n$ is fixed). I can prove that such an upper bound exists  for instance for $n=2$, it's attained by the A2 ("hexagonal") lattice since any lattice has a basis such that the angle between the two basis vectors is between 60 and 120 degrees. I don't know what it is for general $n$. It seems like this should be known, but I can't find it. Does anyone know of a good bound?
There is such a bound due to Minkowski. More precisely he shows for any symmetric convex region C, one can find a basis in xC for sufficiently large x. I cant remember the exact bound offhand, but it is probably somewhere in "An Introduction to the Geometry of Numbers" by Cassels (or any other book on the geometry of numbers). Addendum: on checking I realized that I misremembered Minkowski's result. He does indeed gives a bound for the product of the lengths of a basis in the lattice (in terms of successive minima), but it seems to be a basis for the real vector space rather than a basis for the lattice as you asked for. 


I don't know how good the bound is you can obtain from this, but what about taking a KorkineZolotarev reduced basis of $\Lambda$, say $(b_1, \dots, b_n)$: then, by this paper, $\b_i\_2^2 \le \frac{i + 3}{4} \lambda_i(\Lambda)^2$, where $\lambda_i(\Lambda)$ is the $i$th successive minimum of $\Lambda$. By Minkowski, $\prod_{i=1}^n \lambda_i(\Lambda) \le \gamma_n^{n/2} \det \Lambda = \gamma_n^{n/2}$ (in your case), $\gamma_n$ being the $n$th Hermite constant, whence you get $A \le \prod_{i=1}^n \b_i\_2 \le \frac{\gamma_n^{n/2}}{2^n} \prod_{i=1}^n \sqrt{i + 3}$. 

