Yes, sureley. In Computablilty Theory you have the famous Ackermann-function[1]. It is actually an operator-extension, like you just described. It is important, because it grows overexponentially, but is still computable (unlike e.g. the busy-beaver-function).
Given that Graham's Number is an (over-)extension of the Ackermann-function that means that it's still computable, correct?
Given the series leading up to Graham's Number, G = g_64 (g_1, g_2, ...), BB(n) will outgrow g_n, right? If so what's the smallest n such that BB(n) > g_n?
Graham's number comes from such an extension, and is about 64 such operations deeper. It's useful in the sense that it provides an upper bound for some proof that hasn't yet been proven for ALL numbers.