You could have a separate accumulator for each shuffle, which should allow 28 iterations. (and merge those together at the end of the 28-iteration-loop by widening to u16; at which point you could have an outer loop accumulating in u16 until that runs out)
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For 1, can you raise that to 28 with unsigned accumulators?
14 already assumes unsigned accumulator! 255 [accumulator capacity] / (2 [shuffle-adds] * 9 [highest digit value]) ~= 14
You could have a separate accumulator for each shuffle, which should allow 28 iterations. (and merge those together at the end of the 28-iteration-loop by widening to u16; at which point you could have an outer loop accumulating in u16 until that runs out)