Compute height of AVL tree as efficiently as possible
No.  There's a $\Omega(\lg n)$ lower bound.  You can't do better than $O(\lg n)$ time.  In fact, any algorithm has to visit at least $H-1$ nodes, where $H$ is the height of the tree.

Let $T_1$ be a tree of height $H$, where the balance factor at every node at all but the bottom two levels is 0.  In other words, it is a perfect binary tree of height $H$, except that some subset of the leaves are missing; at the level above the leaves, all nodes have balance factor $+1$, $0$, or $-1$; and at all levels above that, the balance factor is $0$.

Let $T_2$ be an analogous tree of height $H+1$, i.e., it has height $H+1$ and the balance factor at all but the bottom two levels is 0.

Both $T_1$ and $T_2$ are AVL trees.

Now note that any algorithm has to visit at least $H-1$ nodes to distinguish $T_1$ from $T_2$.  Their first $H-2$ levels look identical (every node has two children and has balance factor 0), so you can't tell them apart until you have visited at least $H-1$ nodes.  To count as a correct algorithm for this problem, the algorithm has to produce the right answer on both $T_1$ and on $T_2$.

Consequently, any algorithm has to have worst-case running time at least $H-1$, i.e., $\Omega(H)$.