Is there any formula to calculate how many identical base58 addresses exist in each bit range?
No, you can only estimate that, but you will not get the exact numbers, unless you want to do that in a very small range, where you can just check all of them.
Since each bit is twice the size of it's previous bit, how should I accurately calculate how many of 2^96 collisions could exist in a certain bit range?
Well, if you have 2^256 on one side, and 2^160 on another, you can get 2^(256-160), which is equal to 2^96. It is just size of all public keys, divided by size of all addresses. Nothing more, nothing less, but for such huge numbers, it is only an estimation. It could be 2^95 or 2^97 as well for some specific values, nobody knows the exact number, because nobody counted all of them one-by-one, and nobody found an algorithm, to quickly and accurately count them, without breaking math problems behind them.
Also note that for Script-based hashes, this number is probably much bigger, because then you can have a script, that for example pushes some 520-byte value on the stack.
You mean we can't find an estimation of how many probable identical addresses exist in each bit range?
Estimation is always possible, but it could be wrong, because you could estimate it as 2^96, while it could be 2^97 or more, if you count compressed and uncompressed keys separately, or if you take other things into account.
So it would be wrong to divide N by 2^96 and then divide any bit range by that result?
If you want to get the probability of a collision, then it is something like that.
but is the calculation above correct?
To some extent yes, but you cannot be 100% sure, that the real probability is 5*10^(-29), and not for example 6*10^(-29).
Meaning there are no collisions in 66 bit
It is almost guaranteed that there are collisions in 66 bit. If there would be absolutely no collisions, that would mean RIPEMD-160 is totally broken.
or you are saying we can't really tell because there is no mathematical relation?
There is a mathematical relation, but it is unknown, and hard to simplify, because of the
avalanche effect. However, if you want to for example count, how many collisions are present, where you have some 32-bit hash function on input (for example reduced SHA-256 from 32-bit internal values to 4-bit internal values), and 20-bit hash function on output (for example RIPEMD-160, also reduced to 4-bit internal values), then you can use brute force, to get the exact results.