A limit related to super-root (tetration inverse).
I have some observations, from where someone more experienced might be able to derive the proof - maybe this is helpful)
Let the iterated functional root ("superroot of order") $B(z,n)$ (which finds the " B "ase of the powertower) be defined as
$$ \;^n b = z \qquad \to \qquad B(z,n) = b $$
For the following let us always denote $u$ for the $\log(z)$ for notational convenience.
Let us then define the generalization of the Lambert-W-function to higher iterates as a simple conjugate of the base-finding $B()$-function:
$$ W(u,n) = W(\log(z),n) = \log( B(\exp(u),n)) \tag 1$$
where we can find a formal power series for $W(u,n)$ from the Lagrange series reversion of $u \cdot \exp(u)$, such that
$$ \begin{array}{} \mathcal {\text{ Taylor:} } & W^{-1}(u,1) = u \\
\mathcal {\text{ Taylor:} }& W^{-1}(u,2) = u \cdot \exp(u) \\
\mathcal {\text{ Taylor:} }& W^{-1}(u,3) = u \cdot \exp(u \cdot \exp(u) ) \\
\end{array} \tag 2$$
Then your limit for a general $z$ and $u = \log(z)$ reads
$$ \lim_{n \to \infty} {B(z,n)-z^{1/z} \over \log(z)^n} \qquad \text{or}
\qquad \lim_{n \to \infty} {\exp(W(u,n))-\exp(u \cdot \exp(-u)) \over u^n} \tag 3 $$
Actually we use $z=2$ (and thus $u=\log(2)$) and this reads a bit more friendly:
$$ \lim_{n \to \infty} {\exp(W(u,n))- \sqrt 2 \over u^n} \tag 4$$
Now, if we look at the formal power series for $W(u,n)$ we find a striking pattern, in that with increasing n the coefficients of the equivalently longer leading part become constant, which can be written like the following $$ \begin{array} {} W(u,n) &=& u \Large \left(1-u+{ u^2 \over 2!}-{ u^3 \over 3!} + ... \right. \\ && \qquad + u^{n-1} \Large \left( {a_{n,1} u \over 1!} -{a_{n,2} u^2 \over 2!} + ... \right) \\ && \left. \quad \Large \right)\end{array} \tag 5$$ Let us denote the part in the inner parenthese as $R(u,n)$ $$ R(u,n) = {a_{n,1} u \over 1!} -{a_{n,2} u^2 \over 2!} - ... \tag 6 $$ which makes then $$ \begin{array} {rrl} W(u,n) &=& u \cdot \exp(-u) + u^n \cdot R(u,n) )\\ B(z,n) &=& \exp(u \cdot \exp(-u) + u^n \cdot R(u,n)) \\ &=& \exp(u \cdot \exp(-u)) \cdot \exp( u^n \cdot R(u,n)) \\ &=& \sqrt 2\cdot \exp( u^n \cdot R(u,n)) \\ \end{array} \tag 7$$
Something on convergence: because the first line in fomula (5) contains an exponential series in $u$ and such a series is entire, the consideration of range of convergence of the whole construct focuses on $R(u,n)$ - other than with $z=2$ and $u=\log(2)$ I did not check it, but for this settings it seems to converge to something $\lim_{n \to \infty} R(u,n) =r_\infty \approx 0.0484903140769$ just by numerically searching (binary search) $b_n=B(2,n)$ and computing backwards. Interestingly the coefficients of the power series of $R(u,n)$ have the same scheme to become constant for the leading part which extends when n increases and has thus also a limiting power series. Pari/GP gives me
$$ R(u,12)= u - 5 \cdot u^2/2! + 13 \cdot u^3/3! - 19 \cdot u^4/4! + 1 \cdot u^5/5! + 231 \cdot u^6/6! - ... $$
and this is the same for all values n greater than 12, so we might assume this constant also for the case of $n \to \infty$.
I did not find anything related in OEIS yet, perhaps I can later find some pattern...
We see in the last formula in (7) the squareroot occuring, so we get the initial formula as $$ \lim_{n \to \infty}\sqrt2 \cdot { \exp(u^n R(u,n)) -1 \over u^n} \tag 8 $$ where the exponentialseries (divided by $u^n$) has no constant term and the linear term is $ R(u,n)$ and the following terms are powers of it, multiplied by n'th powers of $u=\log(2) \approx 0.693 \lt 1$.
These are so far only some (organized) observations, a real proof of the existence of a finite limit must be provided by analysis of the power series of $R(u,n)$ (but besides the heuristics I don't have the proof yet)
$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$
Notice the resemblance with the Koenigs function
https://en.m.wikipedia.org/wiki/Koenigs_function
In fact it is a Koenigs function with the variable fixed to the value $2$.
Since $1 < 2 < \exp(1/e)$ and the derivative is not $0$ or $1$ , the Koenigs function converges to the correct value ; your limit.
$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$
This limit is only possible if
$$\lim\limits_{n\to\infty}\frac{\sqrt[n+1]2_s - \sqrt 2}{\sqrt[n]2_s - \sqrt 2}= \ln2$$
To show this , use l'hopital
We get with $f(x) = x^{f(x)}$ :
$ \frac{D x^{f(x)}} {f ' (x)} = \frac{ \sqrt 2 ^2 2\ln(\sqrt2) }{2} = \ln2$.
Qed
This is part of the answer that justifies the RHS of the limit
$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$
With thanks to Tommy1729 for hints.