Number of nodes in Van-Emde-Boas tree of universe size u?
The nice picture in the question does not include the summary nodes nor the min and max fields, both of which are indispensable components of a van Emde Boas tree (vEB tree). A better illustration might be the following picture taken from how to read off the set represented by a van-Emde-Boas tree, which was drawn by Raphael based on a figure in CLRS, where a number in orange is drawn at a min field that marks its presence in the set of integers being represented.


How many nodes are there in the vEB tree implemented like the illustration above?

There are $1 + 5 + 5 \times 3 = 21$ nodes in the illustration above.

For the sake of simplicity, let $w=2^h$ and $u=2^w=2^{2^h}$. A vEB tree over the universe $\{0,1,\cdots, u-1\}$ is of depth $h=\log\log u$.


There is one node of depth 0, which is the root node.   
There are $2^{2^{h-1}} + 1$ nodes of depth 1. 
Each node of depth 1 has $2^{2^{h-2}} + 1$ child nodes of depth 2. 
...
Each node of depth i has $2^{2^{h-i-1}} + 1$ child nodes of depth $i+1$. 
...
Each node of depth $h-1$ has $2^{2^{h-h}}+1=3$ child nodes of depth $h$. 


In total, the number of all nodes is 
$$1 + \sum_{i=1}^{h}\prod_{k=1}^i(2^{2^{h-k}} + 1)=(2^{2^h} - 1)\sum_{i=0}^{h}\frac1{2^{2^i} - 1}\tag{1}$$
which is 1, 4, 21, 358, 92007, 6029862760, 25898063359598159721, $\cdots$ for $h=0,1,2,3,4,5,6,\cdots$ respectively. When $h\ge4$, the number of nodes is about $1.404u$.

Similarly, the number of all summary nodes is, for $h\ge1$,
$$1 + \sum_{i=2}^{h}\prod_{k=1}^{i-1}(2^{2^{h-k}} + 1)=(2^{2^{h}} - 1)\sum_{i=1}^{h}\frac1{2^{2^i} - 1},\tag{2}$$
which is 0, 1, 4, 21, 358, 92007, 6029862760, $\cdots$ for $h = 0,$$1, 2,3,4,5,$$6,\cdots$ respectively. 

What is the number of all non-summary nodes? Subtracting (2) from (1), we obtain

$$(2^{2^{h}} - 1)\left.\frac1{2^{2^i} - 1}\right|_{i=0}=2^{2^h}-1=u-1.$$

The formula also hold for $h=0$. So we have obtained the following surprising formula.
$$\text{the number of non-summary nodes in a vEB tree of universe size } u=2^{2^h}\text{ is }u-1.$$





Every node at the same depth use the same amount of space. Nodes at a smaller depth may use much more space than those at a larger depth. For example, for $h=5$, the root node contains a bit-array of size 65536 but a leaf node just contains several words. Since the number of nodes that use bigger space decreases very quickly as their depth becomes smaller, the total space used is $O(u)$.



Exercise. The number of all leaf nodes, which are vEB trees of universe size 2 in a vEB tree of universe size $u=2^{2^h}$ is $u-1$.