Using well known approximations for the length and number of trailing zeroes of n!, and making the reasonable assumption that the inside zeros appear with frequency $\frac{1}{10}$, we get the following approximation of the total number of zeros, t, in n!:
$t = \lfloor \frac{1}{10}((\frac{\log frac{1}{10}(\frac{\log (2 \Pi n)}{2}+n(\log n)}{2}+n\log (\frac{n}{e})))- \frac{n}{e})- \frac{n}{4}+ \log(n)) + \frac{n}{4} - log(n)\rfloor $
Which simplifies to:
$t = \lfloor \frac{n (9 \ln (10)-4)+4 (n-9) \ln (n)+2 \ln(2 \Pi n)}{40 \ln(10)} \rfloor$
This approximation seems to work well for n up to at least 10,000.
100!, with digit length 158, has less inside zeroes, 6, with 24 trailing, than the normal expectation for a total of 30, with t=36.
98! is "zero-perfect", i.e. inside zeroes appear with exactly frequency $1/10$, with actual total zero count 35 and $t = 35$
Other examples of zero-perfect factorials are: 1009!, 1097!, 1112!, 2993!, 6128!, ....
There appears to be a strong correlation of n having only 0-3 prime factors in {2, 3, 5} if n! is zero-perfect. Uneven n is often a prime number if n! is zero-perfect.
