In 3 dimensions, you can actually count the number of spheres that fit in a rectilinear box.
the optimal packing arrangement has the spheres organized in planar layers, at the vertices of a triangular grid (that is, the centers will e $2r$ apart in one direction, say X, and $\sqrt{3}r$ apart along another direction (say $Y$). The next plane up is staggered so that the sphere centers are directly above the empy spot in the center of the equilateral triangle of sphere centers on the first layer; that places the centers on the second layer p by $\sqrt{8/3}$.
So for all dimensions large compared to $r$, a box of volume $L\cdot W\cdot H r^3$ can hold $\frac{LWH\sqrt{2}}{8}$ spheres (a bit less because of wastage at the edges but we are assuming a large box). Naively dividing by the sphere volume would have given the answer
$\frac{3LWH}{4\pi}$. So the compensation that you estimated as $0.65$ is more like $0.74$.
If you are estimating how many spheres would fit in the box if you just poured them in randomly, without periodic shaking to make them line up optimally, that is a much harder problem, and the estimated space factor of $0.65$ is probably close to reality.
