Write

N = 4 ↑↑↑↑ 4

in ordinary arithmetic.
The first thing to notice is that every stage of the calculation keeps the base 4, so at the very end we again obtain a power of 4:

4 ↑↑ n = 4^(4 ↑↑ (n-1)) , therefore is a power of 4
4 ↑↑↑ n = 4 ↑↑ ( 4 ↑↑↑ (n-1) ) , hence a power of 4
…

Hence

N = 4 ↑↑↑↑ 4 = 4^K = (2^2)^K = 2^M (M = 2K).

Thus the problem is

“For the particular (enormous) exponent M, which decimal digit occurs
 most often in the decimal expansion of 2^M ?”

———————————————————————————————— What is known about the digits of 2^M ? ————————————————————————————————

1. Size.
   M itself is already astronomically large (far beyond a googolplex), so the number of decimal digits of 2^M is

   D = ⌊ M · log10 2 ⌋ + 1 ,
   

again a number that dwarfs every quantity that can be written with ordinary exponential notation.

2. Last few digits.
   By working modulo 10, 100, … one can obtain any fixed finite block of trailing digits of 2^M without expanding the whole number, but that does not give information about how many 0’s, 1’s, …, 9’s occur in the whole expansion.

3. Distribution of the digits.
   It is conjectured that the sequence of digits of the powers of 2 is “normal’’ in base 10, i.e. that in the limit each of the ten digits appears with relative frequency 1/10. But this has never been proved. In fact, even the much weaker statement

   “infinitely many powers of 2 contain more 7’s than 3’s (say)”
   

is still open. Nothing is known that would allow one to tell, for a given exponent M, whether one digit is more frequent than another without essentially writing down the number itself.

———————————————————————————————— Why you cannot avoid computing (almost) all digits ———————————————————————————————— Suppose there were an algorithm that, given M written in binary, told you which digit is most frequent in 2^M while examining only, say, the last 10^{100} digits of 2^M. Then by feeding artificial exponents to that algorithm and looking at its answers one would extract long-range correlations in the decimal expansion of powers of 2; but the existence of such correlations would contradict the widely held (and unresolved) belief that those digits behave pseudo-randomly. No such method is known, and any method capable of deciding the majority digit for an individual power of 2 would represent a breakthrough well beyond the current state of number-theory.

———————————————————————————————— So what can be said? ———————————————————————————————— • Under the unproved “random–digit’’ heuristic one expects every digit to occur about D/10 times, and the excess of the winning digit to be of order √D. In other words, there is no a-priori reason to guess that 0 will beat 7, or 3 will beat 9; each digit should have probability ≈ 1/10 of being the champion.

• From a rigorous standpoint, however, nothing whatsoever is known that would let us decide, for this particular exponent M, which digit wins.

———————————————————————————————— Conclusion ———————————————————————————————— With the mathematics that is available today the only way to discover which digit occurs most often in

4 ↑↑↑↑ 4 = 2^M

is to compute essentially the entire decimal expansion of the number and count. That is hopeless in practice, and any shortcut would require new results about the distribution of the digits of powers of 2—results far beyond what is currently proved. So the honest answer is: we do not know, and no method is known that avoids “calculating the number digit by digit.’’