[Solution 1] We only need to consider the last digit of the calculation result (the first five digits are the same except for (C)), i.e., consider the influence of the last digit of $n$ on the result:
\begin{tabular}{c|cccccccccc}
\hline Last digit of $n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline Last digit of $n^{3}$ & 0 & 1 & 8 & 7 & 4 & 5 & 6 & 3 & 2 & 9 \\
\hline Last digit of $n^{3}-n$ & 0 & 0 & 6 & 4 & 0 & 0 & 0 & 6 & 4 & 0 \\
\hline
\end{tabular}

From the table above, we can see that the last digit of $n^{3}-n$ can only be $0, 4, 6$.
Clearly, the correct answer can only be (A) or (C).
Furthermore, since $n^{3}-n=(n-1) n(n+1)$, it is divisible by 3. A number divisible by 3 has the sum of its digits as a multiple of 3.
Notice that $3+8+8+9+4+4=36$, while
$$3+8+8+9+5+4=37.$$

Therefore, the correct choice is (A).
[Solution 2] Since $n^{3}-n=(n-1) n(n+1)$, then $6 \mid (n^{3}-n)$.
And $6 \mid 388944$,
while $388954=388944+10$, and $388948=388944+4$,
so $6 \nmid 388954, 6 \nmid 388944$.
Also, 388945 is an odd number, so $6 \nmid 388945$.
In summary, the correct result can only be 388944.
Therefore, the correct choice is (A).