I would like to establish an ordering on the hypercubes $[0,1]^n$ for $0 \leq n \leq K$. This means finding a function $V(n): \mathbb{N} \rightarrow \mathbb{R}$ such that we have a strict ordering $V(0) < V(1) < \cdots < V(K)$.
This ordering should
(1) capture my intuition that the cube is bigger than the square, and the square is bigger than the line. For example the cube needs more vertices for its representation than the square.
(2) capture the exponential growth of the hypercubes. For example the sampling density is defined as $M^{1/n}$, with $M$ being the number of samples, n the dimension. To sustain a constant sampling density $M^{1/n}=c$ we therefore need $M = e^{n log(c)}$ samples, which scales exponentially.
Unfortunately, the n-dimensional Lebesgue measure is $V(n) = 1$. The $K$-th dimensional Lebesgue measure is $V(n<K) = 0$ and $V(K)=1$. This does not establish a strict ordering.
I know of three ways to establish a strict ordering on the hypercubes,
(1) computing the longest diagonal in n-dimensions which is $V_1(n) = \sqrt{n}$,
(2) computing the number of vertices of the cube which is $V_2(n) = 2^n$,
(3) discretizing the cube with some resolution $\Delta$, such that each dimension contains $M = 1/\Delta$ points. Then the size will be $V_3(n) = M^n$
The first two ways do not capture the exponential growth of the hypercube, while the third way depends on an arbitrary parameter, such that $V_3$ could either grow too fast or too slow.
Does anyone know of a better way to establish an ordering between unit hypercubes? Is there a generalization of measures for such a case?
EDIT: clarified the "size" issue raised in the comments
The best I have come up with so far: Any n-cube can be decomposed into a set of elements, i.e. a set of m-cubes , for example the cube in 3d is decomposed of 1 3-cube (cell), 6 2-cube (faces), 12 1-cubes (edges) and 8 0-cubes (vertices). As @MvG pointed out in the comments, we can compare cubes by comparing their elements. The number of m-cubes inside an n-cube is given by the formula $$E_{m,n} = 2^{m-n}\frac{n!}{m!(n-m)!}$$ To each m-cube we can assign a value of $m$, such that we can assign to each n-cube the sum of its m-cubes times m as $$V(n) = \sum\limits_{m=0}^n m\cdot E_{m,n}$$
We obtain $$\begin{align*} n \ :\ & 0 < 1 < 2 < 3\ \ < 4\ \ \ < \cdots\\ V(n)\ :\ & 0 < 1 < 6 < 21 < 108 < \cdots \end{align*} $$
This is equivalent to the sequence $a(n)=(n-1)3^{n-2}$ (A027471)