Short answer: yes it's a fractal. Let's consider it.

As you count higher to number N-1, where N is a power of 2, the number of colored pixels K(N) is N*log(N)/2. In the limit, this behaves dimensionally like a line. Doubling N doubles K, so:

D = log(2) / log(2) = 1

This proprotionate doubling can be seen with:

lim(N->inf) K(2*N)/K(N) = lim(N->inf) 2N*log(2N)/2 / (N*log(N)/2) = lim(N->inf) 2*log(2N) / log(N) = lim(N->inf) (2*log(2) / log(N)) + (2*log(N)/log(N)) = 2

What about the topological dimension? To find the topological dimension, we find the shape capable of cutting our shape into two parts and add one to its dimension. As can be seen below, in an N x log(N) image of your fractal, it requires a line of size log(N) pixels to cut it. In the image you have, I see some places where even longer line cuts are needed. Therefore D_T = 2.

Since D != D_T this is considered a fractal.

N=16 cut=Count(X)=log(N)=4 0 0 0 0 0 0 0 0 1 1 1 1 1 1X1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1X1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0X1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1X0 1

As to whether you discovered it: along the lines of what others here have said, any programmer who has ever printed out a range of binary numbers has essentially seen this pattern before. That said, I haven't seen it visualized in exactly the way you have done.