asymptotic of $x^{x^x} = n$

How find the asymptotic behavior for $x(n)$ if $x^{x^x} = n$?

I supposed that $x = O(\log\log{n})$ and took logarithm two times.

So I get $x = O(\frac{\log\log{n}}{\log\log\log{n}})$

Is it right? How to improve this estimate?

Let's write

$$ \begin{align} &\log\log n = L_2(n), \\ &\log\log\log n = L_3(n), \\ &\log\log\log\log n = L_4(n). \end{align} $$

Then from $x^{x^x} = n$ we get

$$ x\log x + L_2(x) = L_2(n), \tag{1} $$

and so, since $x \to \infty$ as $n \to \infty$, we have

$$ x\log x \sim L_2(n) \tag{2} $$

as $n \to \infty$. Taking logs of this yields

$$ \log x + L_2(x) \sim L_3(n), $$

from which it follows that

$$ \log x \sim L_3(n). \tag{3} $$

By dividing $(2)$ by $(3)$ we then find that

$$ x \sim \frac{L_2(n)}{L_3(n)}, \tag{4} $$

agreeing with your estimate. To get the next term we iterate.

Taking logs of $(1)$ and rearranging yields

$$ \log x = L_3(n) - L_2(x) - \log\left(1 + \frac{L_2(x)}{x\log x}\right). \tag{5} $$

From $(3)$ we know that

$$ L_2(x) = L_4(n) + o(L_4(n)), $$

so we can rewrite $(5)$ as

$$ \log x = L_3(n) - L_4(n) + o(L_4(n)). $$

Substituting these into $(1)$ yields

$$ x\Bigl(L_3(n) - L_4(n) + o(L_4(n))\Bigr) = L_2(n) - L_4(n) + o(L_4(n)). \tag{6} $$

Now

$$ \Bigl(L_3(n) - L_4(n) + o(L_4(n))\Bigr)^{-1} = L_3(n)^{-1} + \frac{L_4(n)}{L_3(n)^2} + o\left(\frac{L_4(n)}{L_3(n)^2}\right), $$

so we can conclude from $(6)$ that

$$ x = \frac{L_2(n)}{L_3(n)} + \frac{L_2(n) L_4(n)}{L_3(n)^2} + o\left(\frac{L_2(n) L_4(n)}{L_3(n)^2}\right) $$

as $n \to \infty$.