Let p be a prime number > 5. If p is 1 or 4 in the
modulus of 5, then the Pisano period is a
divisor of p-1. If p is 2 or 3 in the modulus
of 5, then the Pisano period is a divisor
of 2p+2.
Easiest way to estimate that number is to employ the Stirling approximation: you have
$$ n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$
(this is asymptotic notation), or even more roughly speaking, $ n! \approx e^{n\ln n} $ as order of magnitude. In your case, say $ n = 10^{80} $, you have
$$ n! \approx e^{80\cdot\ln{10}\cdot10^{80}} \approx 10^{10^{81.9}}. $$
(EDIT:
I had written 81 instead of 81.9 at the exponent; not only was the rounding incorrect, but especially, rounding exponents needs a lot of care when talking about orders of magnitude!)
By the way, this is also the plausible order of magnitude (in natural units) for the volume of the phase space of the observable universe.
Pictorially speaking, the decimal expansion of this number would need to convert every single particle in the observable universe into a digit in order to be written out.
For an even more pictorial comparison, the age of the universe is thought to be about $ 10^{17} $ seconds. This is far less than the number of digits you would need to write out the above number. Even if you could churn out, say, a billion of digits per seconds, you would still need about $ 10^{82}/10^9 = 10^{73} $ seconds to complete the task, which is a ridiculously prohibitory amount of time.
Best Answer
If you know three things, namely factorisation of $N=\prod_n p_n^{i_n}$,
value of $d(N-1)$
and $m_n$ maximal power of $p_n$ dividing $(N-1)!$ for each $n$ then I think you should be able to calculate $d(N)$.
In that case $(d(N)=d(N-1)/(\prod_n(m_n+1)) \times (\prod_n(m_n+i_n+1)) $ by standard number of divisors formula.
I don't think you can simplfy much more unless I misunderstood your question.