(I'm posting this question in the spirit of this advice and will post an answer if no one else does so.)
[This paper] proves the following:
If $k, x$ and $a_1,a_2,a_3\dots$ are positive integers, then $a_1\uparrow a_2\uparrow\dots a_{s}\uparrow x\pmod k$ is independent of the value of $x$, provided $s\gt h(k).$ Here $h(k):=\min\{n:\lambda^n(k)=1\}$, where $\lambda()$ is the Carmichael function.
Thus, if we're considering the last $d$ decimal digits of such a tower, the proviso involves $k=10^d.$ I've verified by computer that $h(10^d)=d+2$ for $1\le d\le 10^3$.
Q: Is it correct that $\ h(10^d)=d+2\ $ for every positive integer $d$? How to prove it? For other bases $b$ as well, are there correspondingly simple formulas for $\ h(b^d)?$
EDIT: Since my answer would essentially coincide with that of @Onir, let me instead just mention here that the same method gives the following explicit formula for $\lambda^n(10^d), \ \ n,d\in\mathbb{Z_+}:$
$$\lambda^n(10^d)=\begin{cases} 1&\text{if $1\le d\le n-2$}\\ 2&\text{if $d=n-1$}\\ 2^2\,5^{d-n}&\text{if $n\le d\le 2n-1$}\\ 2^{d-2n}\,5^{d-n}&\text{if $d\ge 2n+2$} \end{cases}$$
for $a\geq 4$ and $b\geq 1$ we have
$\lambda(2^a 5^b) = \operatorname{lcm}(2^{a-2},4\cdot 5^{b-1}) = 2^{a-2}5^{b-1}$.
Let $k = \lceil d/2 \rceil$.
Then $\lambda^k(10^d) = 2^25^{d-k}$
From here each application of $\lambda$ just reduces the exponent of $5$ by one, so we have:
$\lambda^d(10^d) = 2^2$
$\lambda^{d+1}(10^d) = 2$
$\lambda^{d+2}(10^d) = 1$