The answer is $1$. Color the cells black and white in a 3D checkerboard pattern, with black in the corners. Then there are $53$ black cells and $52$ white ones. At the signal, every cockchafer moves to a cell with the opposite color, so we will end up with $52$ cockchafers in black cells and $53$ in white ones. Thus, at least one black cell must be empty. To show that this minimum is attained, have the cockchafers in rows 1-2 swap rows, have the remaining cockchafers in columns 1-4 swap columns (1<->2 and 3<->4), and have the remaining cockchafers at height 1-6 swap heights (1<->2, 3<->4, 5<->6). The lone remaining cockchafer at (3, 5, 7) can go wherever.

The answer is $35$. Note that there are $70$ cells in odd-numbered rows and $35$ in row 2. At the signal, every cockchafer moves to a row with opposite parity, so we will end up with $70$ cockchafers in row 2 and $35$ in the odd rows. Hence, at least $35$ cells will be empty. To show that this minimum is attained, move the cockchafers in row 2 according to the following diagram, where blue = move to the corresponding cell in row 1, red = move to the corresponding cell in row 3: [asy][asy] unitsize(1cm); add(grid(7,5,gray)); int[] codes={0, 0, 0, 0, 0, 0, 6, 5, 3, 3, 3, 3, 3, 3, 5, 5, 5, 6, 6, 6, 6, 0, 0, 0, 0, 0, 0, 6, 5, 3, 3, 3, 3, 3, 3}; for (int i=0; i<7; ++i) { for (int j=0; j<5; ++j) { int code = codes[7*j+i]; pen col = red; if (code < 4) { col = blue; } real dx = -1; if (code%4 < 2) { dx = 1; } real dy = -1; if (code%2 == 0) { dy = 1; } draw((i+0.5+0.05*dx, j+0.5+0.05*dy)--(i+0.5+0.95*dx, j+0.5+0.95*dy), col, EndArrow); } } [/asy][/asy] This sends every cockchafer in row 2 to a different cell, so at least $35$ cells in the odd rows will be non-empty. Also, for every cell to which a row 2 cockchafer was sent, have the cockchafer in that cell swap with the row 2 cockchafer, so that all $35$ cells in row 2 are non-empty. The remaining cockchafers can go wherever. This gives at least $70$ non-empty cells, hence at most $35$ empty cells, as desired.