For whoever wondering about the practicality of this algorithm, no, at least for now. Constants aside, the algorithm currently requires an enormous cutoff for the recursion (anything below that should be handled by traditional algorithms):

> Let `n0 := 2^(d^12) >= 2^4096`, and suppose that we wish to multiply integers with n bits. For `n >= n0` we will describe a recursive algorithm that reduces the problem to a collection of multiplication problems of size roughly n^(1/d). We will show that this algorithm achieves `M(n) = O(n log n)`, provided that `d >= 1729`. (p. 33)

2^(1729^12) ~= 10^(2.15 * 10^38). The authors do note the possibility of much lower cutoff:

> In this section we outline a number of such modifications that together reduce the constant to `K = 8 + epsilon`, so that the modified algorithm achieves `M(n) = O(n log n)` for any `d >= 9` (rather than `d >= 1729`). (p. 39)

Here 2^(9^12) has about 85 billion decimal digits. Much smaller, but still too big to be practical.