Notation: ${^n}x = x^{x^{\cdots^x}}$ is tetration, i.e. $x$ to the power of itself $n$ times. $\mathrm{srt}_n(x)$ is the super $n$-th root, or the inverse function of ${^n}x$, which is well defined for $x\ge 1$. I can prove that $$ \lim\limits_{n\rightarrow\infty} \mathrm{srt}_n\left({^{n+1}}2\right) $$ converges to some value between about $\mathrm{srt}_3(256)\approx 2.2915$ and about $2.6$, but it is computationally intractable even for relatively small $n$. For example $^5 2\approx 2\times 10^{19728}$.
For ease of notation, we let $s_n = \mathrm{srt}_n(^{n+1}2)$. I would be very surprised if there's a nice closed form of $\lim\limits_{n\rightarrow\infty} s_n$, so I'm mostly interested here in how to approximate it other than a direct computation of the definition, which really isn't all that viable. As noted above, even computing $s_4$ is tough to do from the formula, though taking some logarithms can get you a bit further, it doesn't help much since tetration is much faster than exponentiation. Is there some trick that convert the formula to $s_n$ into something more tractable?
I can see how you could use the same approach as the one I used (see below) to get better lower bounds than $2.29$, but I suspect that would also become extremely difficult to use if you wanted any sort of precision (even one decimal place might be hard).
Proof of convergence: Clearly $s_n > 2$ for all $n$, so it suffices to show $s_n$ is decreasing. Observe:\begin{eqnarray} ^n s_n &=& 2^{\left(^n2\right)} = 2^{\left(^{n-1}s_{n-1}\right)}<(s_{n-1})^{\left(^{n-1}s_{n-1}\right)} = {^n}(s_{n-1}) \end{eqnarray} since $x\to {^n}x$ is increasing, this implies $s_n$ decreases.
Proof of lower bound: We prove that $s_n > c = \mathrm{srt}_3(256)$ for all $n$ by proving inductively ${^n} c \le\frac{\ln 2}{2\ln(c)} \left({^{n+1}}2\right)$. Since $\frac{\ln 2}{2\ln(c)} <1$, this means ${^n}c<{^{n+1}}2$. Taking $\mathrm{srt}_n$ of both sides shows $s_n>c$.
For the base case, we take $n=2$:\begin{eqnarray} c^{c^c} &=& 256 = 2^8\\ c^c \ln c &=& 8(\ln 2)\\ c^c \ln c &=& \frac12 (\ln 2) 16\\ c^c&=&\frac{\ln 2}{2\ln(c)} \left({^{3}}2\right) \end{eqnarray}
Now, for the inductive step. Suppose ${^n} c \le\frac{\ln 2}{2\ln(c)} \left({^{n+1}}2\right)$ for some $n\ge 2$. Observe that for $x > 4$ (this is not a tight bound):$$ \frac{\ln 2}{2\ln(c)} x < \frac1{\ln c}\ln\left(\frac{\ln 2}{2\ln(c)}\right) + \frac{\ln 2}{\ln c}x $$ Since ${^{n+1}2} > 4$, we therefore have \begin{eqnarray} {^n} c &\le&\frac{\ln 2}{2\ln(c)} \left({^{n+1}}2\right)\\ &<&\frac1{\ln c}\ln\left(\frac{\ln 2}{2\ln(c)}\right) + \frac{\ln 2}{\ln c}\left({^{n+1}2}\right) \end{eqnarray} Taking the $c$th power of both sides yields $$ {^{n+1}} c < \frac{\ln 2}{2\ln(c)} \left(^{n+2}2\right) $$ as desired. Hence we have inductively $$ {^{n}} c < \frac{\ln 2}{2\ln(c)} \left(^{n+1}2\right) $$ for all $n\ge 2$. Therefore $s_n > c$ for all $n$.
Computed with WolframAlpha, the first three terms of $s$ are \begin{eqnarray} s_1 &=& 4\\ s_2 &\approx& 2.74537...\\ s_3 &\approx& 2.58611...\\ s_4 &\approx& 2.57406... \end{eqnarray} Search query used for $s_2$, $s_3$, and $s_4$.