Nothing changes

I'm going to prove that $4$ is minimum. Obviously one 0 is impossible because no matter how many factorials to you "assign" it, you will still get 1. Similarly, if we use two 0's, we get $(0! + 0!) = 2$ , no matter how many factorials to you "assign" it, you will still get 2.

Now if there three 0's, the only possible beginning step is $(0! + 0! + 0!)!$ which is divisible by 6, so no matter how many factorials you assign to it, it will always be divisible by 6. But 80000 is not divisible by 6. So three 0's is impossible too. We are left to show that four 0's is possible.

$\Large {\begin{aligned} && \Bigg [ \bigg[ \quad(0! +0! + 0! + 0!)!! \quad \bigg]!!! \quad \Bigg ]!^{(30)} \\ & = & \Bigg [ \bigg[ \quad 4!! \quad \bigg]!!! \quad \Bigg ]!^{(30)} \\ & = & \Bigg [ [ 2 \times 4 ]!!! \quad \Bigg ]!^{(30)} \\ & = & \Bigg [ \quad 8 !!! \quad \Bigg ]!^{(30)} \\ & = & \Bigg [ \quad 2\times 5 \times 8 \quad \Bigg ]!^{(30)} \\ & = & \Bigg [ \quad 80 \quad \Bigg ]!^{(30)} \\ & = & 20 \times50\times80 = 80000 \end{aligned} }$

My thought process is that basically a mulitfactorial of a number is the product of terms of an arithmetic progression (A.P.). So we just need to find a few A.P terms such that their product is 80000, which is $20\times50\times80$ . Repeat the process again. How do we get 80? It's a product of another few terms of an A.P, in this case $80 = 2\times 5\times 8$ . And how do we get $8$ ? It's $2 \times 4$ .