**Not an answer** Merely remarks.  
Let me use superscript $[k]$ for $k$-fold composition.  $\log^{[3]} n$ means $\log\log\log n$.

As I remarked on the other question, for fixed $a$ and $n$, the value $f(k,a,n)$ becomes complex for large $k$.  Sequence $\log^{[k]}(n)$ decreases as $k$ increases, until it becomes negative, and then complex.  

I have worked with so-called **transseries**.  In that formalism, we are interested in the rate of growth, not the initial part of the function.

<cite authors="Edgar, G. A.">_Edgar, G. A._, [**Transseries for beginners**](http://dx.doi.org/10.14321/realanalexch.35.2.0253), Real Anal. Exch. 35(2009-2010), No. 2, 253-310 (2010). [ZBL1218.41019](https://zbmath.org/?q=an:1218.41019).</cite>

My work on fractional iteration applies only to "exponentiality zero", so does not include $\exp^{[1/2]}$.

<cite authors="Edgar, G. A.">_Edgar, G. A._, [**Fractional iteration of series and transseries**](http://dx.doi.org/10.1090/S0002-9947-2013-05784-4), Trans. Am. Math. Soc. 365, No. 11, 5805-5832 (2013). [ZBL1283.30001](https://zbmath.org/?q=an:1283.30001).</cite>

----

Some remarks on the question here.  Is there some reason to use exponentiation base $2$ and not $e$?  I will use exponential and logarithm base $e$ instead of base $2$.  Let me shift your index $k$ by $1$.  Then define
\begin{align}
f(0,a,n) &:= an,
\\
f(1,a,n) &:= \exp f(0,a,\log n) = \exp(a\log n) = n^a
\\
f(2,a,n) &:= \exp f(1,a,\log n) = \exp(\exp(a \log(\log n)))
=\exp((\log n)^a)
\\ &\qquad \dots
\\
f(k,a,n) &:= \exp^{[k]}(a\log^{[k]}(n)) = \exp^{k-1}\Big(\big(\log^{[k-1]}(n)\big)^a\Big)
\end{align}
Write $M_a$ for the multiplication function: $M_a(n) = an$.  
We are interested in a "limit" in some sense as $k \to \infty$.
I doubt that it converges according to the natural topologies
for transseries.  

Note that
\begin{align}
f(k,a,n) &= \exp^{[k]}\circ M_a\circ \log^{[k]} (n)
\\
f(k,a,f(k,a,n)) &=\exp^{[k]}\circ M_a\circ \log^{[k]}\circ \exp^{[k]}\circ M_a\circ \log^{[k]} (n)
\\ &= \exp^{[k]}\circ M_a\circ M_a\circ \log^{[k]} (n)
\\ &= \exp^{[k]}\circ M_{a^2}\circ \log^{[k]} (n)
\\ &= f(k,a^2,n)
\end{align}
So if $g(a,n) = \lim_{k \to \infty} f(k,a,n)$ in some sense, then
we might expect that
$$
g(a,g(a,n)) = g(a^2,n) .
$$
So: prove by induction on $k$ that $f(k,a,n) < e^n$.  Then conclude that $g(a,n) < e^n$, and $g(a,g(a,n)) = g(a^2,n) < e^n$ (good for what we want).
But also
$$
g(a,g(a,g(a,n))) = g(a^3,n) < e^n
\\
g(a,g(a,g(a,g(a,n)))) = g(a^4,n) < e^n
$$
So that second iterate $g(a,g(a,n))$ is much, much slower than $e^n$.