There's two answers here: a real answer, and a (slightly) cheating answer.
6 is the largest number where it represents the number of segments.
First, we can put an upper bound on our answer. At most, we have a two digit number. This is because the smallest three digit number (100) is an order of magnitude more than the maximum number of displayed segments ($21 = 7\times3$).
Through a process of elimination, we can work backwards from the largest possible 2 digit number: 14 ($=7\times2$).
We end up with $6$ being the largest number.
If we can front-pad the number with zeros, we can get to infinity, but only certain numbers. This is because each zero we add in front of the number is 6 additional segments.
For example, $0...0400$ can be made with 114 leading zeros. $(114+2)\times6+4 = 400$.
In fact, there's a limited set of numbers that answer this. We know these numbers are greater than six, and we know their remainder when divided by six is equal to the sum of their segments mod 6.
Let $N$ be the number, and $s_n$ be the number of segments of the $n$th digit.
$$\left(\sum s_n\right) \mod 6 = N\mod6$$
Any number of the form $0...06...6$ will work (assuming enough zeros). Also, any number of the form $0...040...0$.
Some things I've worked out:
- $N$ is even, iff $\sum s_n$ is even. This is because zero has an even number of segments, and you can't add even numbers together to get odd numbers. The corollary to this is: $N$ is odd, iff $\sum s_n$ is odd. This may seem really obvious, but it helps tremendously.
- 1, 2, 8, and 9 are "backwards" numbers. By this, I mean that they are even, but their number of segments is odd, or vice versa. This means that a number cannot only be made of 1s or 9s, and a number made solely of 2s and/or 8s must have an even number of 2s and 8s. A corollary of this is that a number made of solely "regular" odd numbers cannot have an even number of digits. (These numbers being 3, 5, and 7)
There are probably many other forms. I'd be interested to see a mathematician work through them all.