I think VSJ is right about the limits but wrong about the number of ordered partitions.

To work on this problem, it is helpful to have a method to generate a random ordered partition. This is very easy - flip $n$ coins, and divide them into streaks. So if the flips are HTTHHHTHTTTTH, the corresponding ordered partition is $1+2+3+1+4+1$, and the corresponding unordered partition is $1+1+1+2+3+4$. Since each ordered partition is equally likely to occur (it comes from exactly two sequences of coin flips), this question is equivalent to the question: Which unordered partition is most likely to occur, and what is its probability of occurrence?

The law of large numbers suggests that the most likely unordered partition is exactly what VSJ said - the maximum likelihood $a_i$ should be close to the expected value of $a_i$, which is $n/2^{i+1}$.

But since the probability of any individual unordered partition occurring, even the most likely one, is clearly $o(1)$, the number of occurrence of any unordered partition is $o(2^n)$, and thus can't be $c2^n$.

More precisely, the Central limit Theorem shows that the probability of getting a given $a_1,a_2,\dots, a_k$ is $O(1/n^{k/2})$, so the number of ordered partitions corresponding to a given full unordered partition must be $o(2^n/n^{k/2})$ for all $k$.