A function that grows faster than any function in the sequence $e^x, e^{e^x}, e^{e^{e^x}}$...

a negative answer
No, there is no such function constructed in a "natural" way, if we construe that word propertly.

Perhaps consult the literature on "transseries" ...

For example, transseries in the sense here: G. A. Edgar, "Transseries for Beginners". Real Analysis Exchange 35 (2010) 253-310 .

The set of transseries includes real elementary functions, is closed under indefinite integration, composition, and many other operations.

But no transseries has growth rate beyond all $e^{\dots e^x}$. There is an integer "exponentiality" associated with each (large, positive) transseries; for example Exercise 4.10 in: J. van der Hoeven, Transseries and Real Differential Algebra (LNM 1888) (Springer 2006)

A (large, positive) transseries with exponentiality $n+1$ grows faster than any transseries with exponentiality ${} \le n$. And $e^{\dots e^x}$ with $n$ exponentiations has exponentiality $n$.

Here is a proof that there is no such function und elementary function.

Let $A$ be a collection of functions which grow slower than the collection $e^x, e^{e^x} , \dots $.

We know that the sum, product composition and integral of functions which grow slower than $e^x, e^{e^x} , \dots $ also grows slower that $e^x, e^{e^x} , \dots $.

Then the set of functions which are finite iterate sums, $\dots$ of functions in $A$ also grows slower that $e^x, e^{e^x} , \dots $

Now, depending on or definition of elementary function, if you take $A= \{1, x, \exp, \log, x^a \}$ you should be done.