Now that you know the answer (it reverses the deck), here's how you can prove it. The shuffle can be written as a cyclic permutation. If I write $$( 1 \ 2 \ 4 \ 8 \ 7 \ 5 )(3 \ 6),$$ this is a way of writing the permutation in which the first card goes to the second place, the second goes to the fourth place, the fourth goes to the eighth place, etc. This is just a different way of writing the result of one inshuffle. Now iterating this permutation three times looks like this: $$( 1 \ 2 \ 4 \ 8 \ 7 \ 5 )(3 \ 6)( 1 \ 2 \ 4 \ 8 \ 7 \ 5 )(3 \ 6)( 1 \ 2 \ 4 \ 8 \ 7 \ 5 )(3 \ 6),$$ which reduces to the cycle $$(1 \ 8)(2 \ 7)(3 \ 6)(4 \ 5),$$ which is exactly the permutation which reverses the order.
Maybe you can try it for 16 and observe if there is a nice pattern which works for all $2^n$.