As in the article, let
X = # sixes in N rolls
The event of interest is:
{X >= N/6}
It turns out this is a decreasing function of N, and that fact answers the question ("A is most likely"). But why does it decrease in N?
As you say, the distribution of X flattens out as N increases. The probability mass is spread out over the numbers 0...N, with a peak at exactly N/6. (Not N/6 - 1, or N/6 + 1, but N/6.)
About half the probability mass is between 0...N/6, and half is between N/6...N. So in general, the probability we care about (A, B, or C) is rather close to 1/2 (because it is just the right half of the distribution).
Since the distribution is flattening out, as N increases, mass moves out of the exact center (X = N/6) and toward the left or right. Since the event of interest is
{X >= N/6}
{X > N/6},
In short: Because the event of interest contains the very center, as more rolls are done, the overall distribution flattens, and the interval between [N/6...N] benefits less and less by containing the (ever-more-diluted) center point.
This is another way to restate Stigler's explanation at the top-right corner of page 401 in the OP. ("The ranking ... reflects the fact that ... P(X = Np) decreases as N increases and the distribution spreads out.")