I read some articles about the fast growing hierarchy (and saw some vids), and I wonder how to calculate:
I know that there are huge numbers, but I wonder if someone here can give me estimate how big are those huge numbers...
i.e. for example - how many digits there are at each one of them.
On
It depends on the meaning of calculate.
If you want a decimal string representation, then it is not really possible, as the number of bits required to store that string would very soon exceed the number of atoms in the universe.
On the other hand, you can prove various properties of these numbers. For example, what are the last three digits of $f_4(4)$? How does it compare to other big numbers, such as those given by Knuth's up-arrow notations? etc.
It is the same philosophy as what we mean by "calculating $\pi$" or "calculating $\sqrt 2$". You are never able to represent its decimal expansion with finite amount of information, but this doesn't stop you from proving results such as $\frac 1 {1^2} + \frac 1 {2^2} + \cdots = \frac {\pi^2} 6$.
One can directly compute $f_3(3)$ using the definition to get
\begin{align}f_3(3)&=f_2(f_2(f_2(3)))\\&=f_2(f_2(24))\\&=f_2(402653184)\\&=402653184\times2^{402653184}\end{align}
Taking the base 10 log of this gives
\begin{align}\log_{10}(402653184\times2^{402653184})&=\log_{10}(402653184)+402653184\log_{10}(2)\\&\simeq121210694\end{align}
which is the amount of digits this has.
$f_4(4)$ cannot be worked out similarly. The closest representable expression for the amount of digits $f_4(4)$ has is $f_4(4)$. A rough lower bound can be obtained by expanding it out and using $f_3(n)>2^n\uparrow\uparrow n\ge16\uparrow\uparrow n$ for $n\ge4$.
\begin{align}f_4(4)&=f_3(f_3(f_3(f_3(4))))\\&>16\uparrow\uparrow(16\uparrow\uparrow(16\uparrow\uparrow(16\uparrow\uparrow4)))\\&>4\uparrow\uparrow\uparrow5\end{align}