This is a nice question. The fission diagram forms a tree, and the leaves of the tree must sum to $n$. By noting that $3$ is the smallest number that can be split, we see that there can be at most $n/3$ ones, and hence there are at most $\frac{2}{3}n$ leaves. Each vertex that is not the base point or a leaf must have degree at least $3$, that is each node that is not a leaf must split into at least two more nodes, so we can bound the total number of nodes by $$F(n)\leq \frac{2n}{3}+\frac{2n}{6}+\frac{2n}{12}+\frac{2n}{24}+\cdots =\frac{4n}{3}$$ which gives a linear upper bound on the number of nodes.