Bound the sum of leaf depth on a complete binary tree of $n$ leaves

A complete binary tree is defined as a tree where each node has either 2 or 0 children.

For a complete binary tree with $$n$$ leaves, there can be different arrangements of nodes, let's define the maximum depth of such a tree as $$d$$, it's clear that $$d \ge \log_2n$$.

The quantity I'm interested in is, the sum of depth of all leaves: $$D$$. What's the upper and lower bound of $$D$$ given $$n$$ and $$d$$?

When you construct such a tree and to add two more nodes as offspring to a leaf at depth $$x$$, what happens to $$D$$ is you subtract $$x$$ and add $$2(x+1)$$, therefore $$D_{new} = D_{old} + x + 2$$.
From this you can see that the best strategy to maximize $$D$$ is to always add nodes to the lowest node (furthest from root), therefore obtaining a tree that has one leaf on every level (except the 0th where the root is) and two leaves at the lowest level. The depth of this tree will be $$(n+1)/2$$ ($$n$$ is always odd for these trees). Therefore $$D \le \frac{n+1}{2}-1+\sum_{i=1}^{\frac{n+1}{2}-1}i=\frac{n^2+4n-5}{8}$$ is your upper bound.
On the other hand if you want to minimize $$D$$ you want to always add children nodes to be closest to the root. Then your depth is $$\log_2(n+1)$$ (the last level is one less than depth) and you have again $$(n-1)/2$$ leaves. Therefore the lower bound is $$D \ge\frac{n-1}{2}\cdot\log_2(n+1)-1$$ for trees with $$2^k -1$$ nodes. For nodes with different number of leaves the bound still holds, but it won't be equal.