For prime $m=p$ you have Lucas' theorem. It might not be obvious, but this gives a fractal structure to the binomial coefficients, modulo $p$.

(From here out, $m$ is an arbitrary integer, not $p$, because this is the statement from Wikipedia.)

It says that if $m=\sum_{i=0}^k m_i p^i$ and $n=\sum_{i=0}^k n_ip_i$ are the base $p$ representations (so $0\leq m_i,n_i<p$) then:

$$\binom{m}{n}\equiv \prod_{i=0}^k \binom{m_i}{n_i} \pmod p$$

Specifically, then, this means that if $0\leq a< b<p^k$ and and $M,N$ are any non-negative integers, then:

$$\binom{Mp^k + a}{Np^k+b} \equiv 0\pmod p$$

This gives big blocks where Pascal's triangle is zero, as in Sierpinski.

More generally, we might want to ask about modulo prime powers.

This answer gives some details on Andrew Granville's result for modulo prime powers. Don't know how "fractal" that result is. It seems to be not as "fractal" is the result
for primes, but it is hard to tell.

If it is, then modulo any number $M$, Chinese Remainder Theorem says that Pascal's triangle looks like the overlay of the fractals for each of the prime powers in the unique factorization.

Without Granville, if $M$ is square-free, there will be a fractal structure, modulo $M$.