Here is a proof that $s(2)=3$. Form a graph where the faces are bricks, the the edges are the boundaries of bricks, and the vertices are places where two bricks intersect. Suppose that no brick is a hexagon or larger. Then the number of edges in a large reason is no more than $5/2$ the number of faces, and the number of vertices is exactly $2/3$ the number of edges, so the Euler number is at least $F -5/2(1-2/3)F=F/6$ which is $O$ of the area of the region, where it should be $O$ of the boundary. Or "the graph is somewhere between a cube and a dodecahedron, but nowhere near an infinite plane"
Therefore, some face has at least 6 edges. Each edge has length and least $1$, since the two vertices can share at most two faces, so the other faces at each vertex are nonadjacent, so have distance at least $1$. Therefore the perimeter of some face is at least $6$. The perimeter of $[0,a]\times [0,b]$ is of course $2(a+b)$, so $s(2)\geq 3$. There is an explicit example with $s(T)=3$, so we are done.
Obviously parts of this argument generalize to higher dimensions, but it is not clear to me if one can patch up the other parts to make it usable.