$\epsilon$-nets with respect to the cut norm
Here is an easy estimate.  Here we call a set S⊆X an ε-net of a metric space X when for every point x∈X, there exists a point s∈S such that the distance between x and s is at most ε.  If you want a strict inequality in the definition of ε-net, you can tweak the value of ε slightly.

It holds that ||A||∞ ≤ ||A||C ≤ n2||A||∞, where ||A||∞ denotes the entrywise max-norm of an n×n matrix A.

It is easy to construct an ε-net of the metric space ([0,1]N, d∞) with size ⌈1/(2ε)⌉N, and it is not hard to show that this size is the minimum.  (To show the minimality, consider the ⌈1/(2ε)⌉N points whose coordinates are multiples of 1/⌈1/(2ε)−1⌉ and show that the distance between any two of these points is greater than 2ε.)  By setting N=n2 and combining this with the aforementioned comparison between the cut norm and the max-norm, the minimum cardinality of an ε-net with respect to the cut norm is at least ⌈1/(2ε)⌉n2 and at most ⌈n2/(2ε)⌉n2.



Update: If my calculation is correct, a better lower bound Ω(n/ε)n2 can be obtained by the volume argument.  To do this, we need an upper bound on the volume of an ε-ball with respect to the cut norm.

First we consider the “cut norm” of a single vector, which is the maximum between the sum of positive elements and the negated sum of negative elements.  It is not hard to show that the volume of an ε-ball in ℝn with respect to this “cut norm” is equal to

$$ \varepsilon^n \sum_{I \subseteq \{1,\dotsc,n\}} \frac{1}{|I|!} \cdot \frac{1}{(n-|I|)!}
= \varepsilon^n \sum_{r=0}^n \binom{n}{r} \frac{1}{r!(n-r)!} $$

$$ = \frac{\varepsilon^n}{n!} \sum_{r=0}^n \binom{n}{r}^2
= \frac{\varepsilon^n}{n!} \binom{2n}{n}
= \frac{(2n)!\varepsilon^n}{(n!)^3}. $$

Next, since the cut norm of an n×n matrix A is greater than or equal to the cut norm of each row, the volume of an ε-ball in ℝn×n is at most the nth power of the volume of an ε-ball in ℝn.  Therefore the size of an ε-net of [0,1]n×n must be at least

$$ \frac{(n!)^{3n}}{(2n)!^n \varepsilon^{n^2}} = \left(\Omega\left(\frac{n}{\varepsilon}\right)\right)^{n^2}, $$

where the last equality is a boring calculation in which we use Stirling’s formula: ln n! = n ln n − n + O(log n).