I can prove that $area(M)\geq 3d^2/8$. First of all find $v\in \mathbb Z^2$ such that $w_v(M)$ is minimal. Than by an affine transform $a$ put $v$ into vector $(1,1)$.
Now draw lines which give us widths of $a(M)$ in the directions $(1,0),(0,1)$. Now $a(M)$ is inside a horizontal stripe of width at least $d$, inside a vertical stripe of width at least $d$ and inside a diagonal (in the direction $(1,-1)$) stripe of width exactly $d/{\sqrt{2}}$.
Here I use usual widths.
Now look at the picture - we see a 6-gon (or 3,4-gon) $B$ as the intersection of the strips, we know that on each side of $B$ there is a vertex of $a(M)$. By playing with vertices one can find all the extremal cases (so, vertices of $a(M)$ must coincide with vertices of $B$).
Finally, in all the extremal cases the area of $a(M)$ is no less than $3d^2/8$