Number of ways to label die with opposing faces summing to 7

Note that the faces with labels 1, 2, and 3 must appear in a 3-cycle (share a corner), and labeling these three faces uniquely determines the labels of the other three faces. There are two ways to do that.

Your successive face-swapping doesn't produce as many unique labelings as you think. Once you swap faces twice, you are back in the original orientation, as this figure shows:

The large numbers are on the faces, and the small ones are on the hidden faces that they are nearest. Swapping the top/bottom and left/right pairs produces the same labeling as simply rotating 180° around the front-back axis.

So here is another way to go about counting the labelings: Fix a single corner of the cube and consider the three faces incident on that corner:

• There are six possibilities for the first face: any number from 1 to 6;
• Once the first face is chosen, there are four possibilities for the second face, because neither the number on the first face, nor its complement, which has to be placed opposite the first face, are available.
• Once the first two faces are chosen, there are only two possibilities for the last face.

So there are $6 \times 4 \times 2$ ways to label the cube if it's in a fixed position. Of course, we are allowed to rotate the cube, and labelings of the cube should be considered equivalent if one can be rotated into the other. There are $4!$ rotations of the cube. So the number of distinct labelings modulo rotation is $$ \frac{6 \times 4 \times 2}{4 \times 3 \times 2 \times 1} = 2 $$