Again not an answer, but just some more context and evidence that this problem seems hard. As Aaron Meyerwitz points out this is typically called the *refinement* order and seems like a difficult problem. Let $P_n$ be the poset of integer partitions partially order by refinement. This poset has minimal element $\hat{0} = 1^n$ the partition with all parts $1$ and maximal element $\hat{1} = n$ the partition with a unique part $n$. The quantity you are looks for is
$$F(n) = \sum_{\lambda \in P_n} |[\hat{0}, \lambda]|$$
the sum of the sizes of all lower order ideals in $P_n$.

Zeigler has a paper [On the Poset of Partitions of an Integer][1]. Essentially this paper shows that the poset $P_n$ lacks some nice properties and is rather unpredictable.

The poset $P_n$ is ranked where the elements at rank $k$ are partitions with $n-k$ parts. So the Whitney numbers of the second kind are $p(n,n-k)$ the number of ways to partitions $n$ into $n-k$ parts. There is a recurrence for $p(n,k)$ (see for example the pretty well developed MathWorld page on [partitions][2]). However, I see a problem in the fact the intervals at the same rank can be very different. For example consider $[\hat{0},2^n]$ and $[\hat{0}, (n+1,1^{n-1})]$ in $P_{2n}$. We have $|[\hat{0}, 2^n]| = n$ since it is just a chains with elements $(2^k, 1^{n-2k})$. On the other hand $|[\hat{0}, (n+1,1^{n-1})]| = p(n+1)$ the number of partitions of $n+1$ which has the asymptotic formula $p(n) \sim exp(\pi \sqrt{2n/3})/(4n\sqrt{3}))$ which is due to Hardy and Ramanujan.

At this time I only see trivial and crude bounds for $F(n)$ is terms a $p(n)$. Though I would be happy to seem someone more knowledge and clever than me produce a nice bound.


 [1]: http://www.mi.fu-berlin.de/math/groups/discgeom/ziegler/Preprintfiles/001PREPRINT.pdf
 [2]: http://mathworld.wolfram.com/PartitionFunctionP.html