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. 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). 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.

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. 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). 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 with more knowledge and more clever than me produce a nice bound.