Limit of a recursively defined bivariate function.  Let m and n be positive integers.
Let $f(m,0)=m$
Let $f(m,n)= e \ln(f(m,n-1))$
$$\lim_{m\to\infty} \ln(m)\Big(f(m,\lfloor\ln m\rfloor)) - e\Big) = 163^{1/3}+C$$
Where $C$ is a constant.
It seems $0.005 > C > 0$
Is this true ? Why is this so ? Is $C = 0$ ?
Is this an analogue to the computation of the Paris constant ?
Can we give a closed form for $C$ ?
EDIT : 
Conjecture : 
$$\lim_{m\to\infty} \ln(m)\Big(f(m,\lfloor\ln m\rfloor)) - e\Big) = A$$
$A > 0$
Is this true ? How to prove this ?
 A: This is not true.  For any $x$, eventually $f(n)$ will fail to be defined (and for any reasonable sized $x$, rather quickly).  For $x=10^{23}$ it only takes five steps.  $\lim_{n \to \infty} f(x)$ does not exist.
Added:  Your notation is confusing.  It would be better to give $f$ arguments of $n$ and $x$, so $f(0,x)=x, f(n,x)=e \ln (f(n-1,x)$, which I will use.  I still don't believe the result with the $e$ multiplying.  If $f(n,x)=e+\delta$, (thinking of $\delta$ as small), $f(n+1,x)\approx e+\delta-\frac {\delta^2}{2e^2}$ so $\lim_{n \to \infty}f(n,x)$ must be the fixed point of your function, which is $e$.
A: Let $h(x)=e\ln x$. If $x>e$ then $h(x)>e$. On the other hand, $h(x)\ge x$ iff $\frac{\ln x}x\ge\frac1e$, and $\frac{d}{dx}\left(\frac{\ln x}x\right)=\frac{1-\ln x}{x^2}<0$ for $x>e$, so $h(x)<x$ for $x>e$. Thus, for any $x>e$ the sequence $\langle h^n(x):n\in\Bbb N\rangle$ is decreasing and bounded below by $e$, so it converges. (Here $h^n$ is the $n$-th iteration of $h$.) Its limit $g(x)$ must be a fixed point of $h$. 
Now $\frac{d}{dx}(x-e\ln x)=1-\frac{e}x$, so $x-f(x)$ has a minimum at $x=e$, where it is $0$. In other words, $x=e$ is the only fixed point of $h$. In other words, $g(x)=e$ for all $x\ge e$, and $\lim_{x\to\infty}\frac1{g(x)-e}$ does not exist.
A: Write $k(m) = \ln(m)\big(f(m,\lfloor \ln(m)\rfloor)-e\big)$.  Here is a graph of $k\big(10^{\displaystyle 10^x}\big)$ according to Maple.  The horizontal line is $163^{1/3}$.  

I see no reason to think $163^{1/3}$ has anything to do with the limit.
A: It is intresting to note that if we replace (in the limit) $m$ by $m+O(m^{\frac{1}{2}})$ we would arrive at the same value. This leads to the simplification in the limit : replace both $log(m)$ by $m$.
This is probably the first step.
The second step is probably finding a good taylor series (converging for the desired interval) and error term for $e$ $ log(m)$ so that the rate of $f(m,m)$ can be expressed. 
This problem them resembles many other problems concerning iterations and limits and should thus be solvable. I note that $exp(\dfrac{x}{e})$ has a parabolic fixpoint. 
Not a solution yet but worth it imho. 
A: In case this helps (too long for comment): Letting $h_n(x) = f_n(x)/e$ we get $h_1(x) = 1 +\log(x)$, $h_2(x) = 1 + \log(1 + \log(x))$, etc- 
So $h_n(x) = L^{[n]}(x)$ where $L^{[n]}$ is the function $1 +\log(\cdot)$  composed with itself $n$ times. Clearly $L^{\infty}(x)= 1$ for any $x$.
But what the OP wants (I'm starting to suspect) is to find 
$$\lim_{x\to \infty} L ^{\lfloor \log(x) \rfloor} (x)$$
