Shortest terminating program whose output size exceeds Graham's number

Haskell, _{59 57 55} 63

(f%s)1=s;(f%s)n=f.(f%s)$n-1
main=print$((flip((%3)%(3^))3)%4)66

How it works: % simply takes a function and composes it n-1 times on top of s; i.e. %3 takes a function f and returns a function of n that equals applying it f to 3, n-1 times in a row. If we iterate the application of this higher-order function, we get a fast-growing sequence of functions – starting with exponentiation, it's exactly the sequence of Knuth-arrow-forest sizes:
((%3)%(3^))1 n = (3^)n     = 3ⁿ = 3↑n
((%3)%(3^))2 n = ((3^)%3)n = (3↑)ⁿ⁻¹ $ 3 = 3↑↑n
((%3)%(3^))3 n = (((3^)%3)%3)n = (3↑↑)ⁿ⁻¹ $ 3  = 3↑↑↑n
and so on. ((%3)%(3^))n 3 is 3 ↑ⁿ 3, which is what appears in the calculation to Graham's number. All that's left to do is composing the function (\n -> 3 ↑ⁿ 3) ≡ flip((%3)%(3^))3 more than 64 times, on top of 4 (the number of arrows the calculation starts with), to get a number larger than Graham's number. It's obvious that the logarithm (what a lamely slow function that is!) of g₆₅ is still larger than g₆₄=G, so if we print that number the output length exceeds G.

⬛

GolfScript (49 47 chars)

4.,{\):i\.0={.0+.({<}+??\((\+.@<i*\+}{(;}if.}do

See Lifetime of a worm for lots of explanation. In short, this prints a number greater than f_ω^ω(2).

Pyth, 29 28 bytes

M?*GHgtGtgGtH^ThH=ZTV99=gZTZ

Defines a lambda for hyper-operation and recursively calls it. Like the definition for Graham's number, but with larger values.

This:

M?*GHgtGtgGtH^3hH

Defines a lambda, roughly equal to the python

g = lambda G, H:
  g(G-1, g(G, H-1)-1) if G*H else 3^(H+1)

This gives the hyper-operation function, g(G,H) = 3↑^G+1(H+1).
So, for example, g(1,2)=3↑²3 = 7,625,597,484,987, which you can test here.

V<x><y> starts a loop that executes the body, y, x times.
=gZT is the body of the loop here, which is equivalent to Z=g(Z,10)

The code:

M?*GHgtGtgGtH^3hH=Z3V64=gZ2)Z

Should recursively call the hyperoperation above 64 times, giving Graham's Number.

In my answer, however, I've replaced the single digits with T, which is initialized to 10, and increased the depth of recursion to 99. Using Graham Array Notation, Graham's Number is [3,3,4,64], and my program outputs the larger [10,11,11,99]. I've also removed the ) that closes the loop to save one byte, so it prints each successive value in the 99 iterations.