Last days, I bothered to find some function f(n) that will be on par with hypothetically definable \(f_{\omega_1}(n)\). By the definition \(\omega_1\) can't has any fundamental sequence. I propose the following way to determine an order type (called: \(\beta\)) for the function f(n):
Define f-typed complexity of ordinal \(\alpha\) as the smallest natural n such that \(f_\alpha(n) < f(n)\). If it doesn't exists, then the complexity is 0. Then the n-th term of fundamental sequence for the ordinal \(\beta\) will be as follows:
If any ordinals with complexity n exist, then the term is the largest of them.
If none of them exist, then the term is the previous term+1.
So, for any f(n) we can construct the countable ordinal (as it has a fundamental sequence) measuring its order type. It can lead to somewhat strange result: I(n) can't diagonalize over all countable ordinals. Even if we define E(n) to be the largest number expressable with n English words (and make it well-defined eliminating inconsistent definitions), still: the ordinals will be countable.
Update 02.08.2013[]
Suppose there is a function f(n) which outgrows each countable ordinal in FGH. For each n there will be an ordinal \(\alpha_n\) so that \(f_{\alpha_n}(n)>f(n)\). So, there is a question: how big is a gap between two consecutive ordinals: \(\alpha_m\) and \(\alpha_{m+1}\). Since they are all countable, then this gap will be countable too, and we can have only countable number of this ordinals (actually: n of them).