How can I find this limit involving thrice-iterated logarithm?

You can find it with a lot of patience ! I give you what I did (hoping that a simpler solution will be provided) :

First, I look at the exponent and, going first to logarithms, obtain $$\frac{(x+1)^{\frac{1}{x}}}{x}=\frac{e}{x}-\frac{e}{2}+\frac{11 e x}{24}-\frac{7 e x^2}{16}+O\left(x^3\right)$$ Then $$(x+1)^{\frac{(x+1)^{\frac{1}{x}}}{x}}=e^e-e^{1+e} x+\frac{1}{24} e^{1+e} (25+12 e) x^2+O\left(x^3\right)$$ So$$A=x+(x+1)^{\frac{(x+1)^{\frac{1}{x}}}{x}}=e^e+\left(1-e^{1+e}\right) x+\frac{1}{24} e^{1+e} (25+12 e) x^2+O\left(x^3\right)$$ Now, let me play with the logarithms $$\log A=e+\left(e^{-e}-e\right) x+\left(\frac{25 e}{24}+e^{1-e}-\frac{e^{-2 e}}{2}\right) x^2+O\left(x^3\right)$$ $$\log\log A=1+\left(e^{-1-e}-1\right) x+\left(\frac{13}{24}-\frac{1}{2} e^{-2-2 e}-\frac{1}{2} e^{-1-2 e}+e^{-1-e}+e^{-e}\right) x^2+O\left(x^3\right)$$ $$\log\log\log A=\left(e^{-1-e}-1\right) x+\frac{1}{24} e^{-2-2 e} \left(-24-12 e+48 e^{1+e}+24 e^{2+e}+e^{2+2 e}\right) x^2+O\left(x^3\right)$$ where you can notice that the first term is $$-x\left[1-\frac{1}{e^{e+1}}\right]$$ So, the limit is $$\frac{1}{24} e^{-2-2 e} \left(-24-12 e+48 e^{1+e}+24 e^{2+e}+e^{2+2 e}\right)=\frac{1}{24}+\frac{1}{2} e^{-2 (1+e)} (2+e) \left(2 e^{1+e}-1\right)$$

All of the above used successively the development (Taylor series) of $\log(1+y)$ close to $y=0$.

Let's give it a try using elementary techniques. First we take care of $f(x) = (1 + x)^{(1 + x)^{1/x}/x}$. Clearly $$\log f(x) = (1 + x)^{1/x}\frac{\log(1 + x)}{x} \to e\tag{1}$$ so that $f(x) \to e^{e}$. We have to deal with the expression $$F(x) = \dfrac{\log \log \log(x + f(x)) + x(1 - e^{-e - 1})}{x^{2}}\tag{2}$$ It can be seen that both the numerator and denominator of $F(x)$ tend to $0$. I am somehow forced now to use L'Hospital Rule. This gives us another complicated expression $$G(x) = \frac{1}{2x}\left\{\left(1 - e^{-e - 1}\right) + \frac{1}{\log \log (x + f(x))}\cdot\frac{1}{\log(x + f(x))}\cdot\frac{1 + f'(x)}{x + f(x)}\right\}\tag{3}$$ The challenge now is to calculate $f'(x)$ and show that it tends to $-e^{e + 1}$ as $x \to 0$. This will ensure that we can apply LHR on $G(x)$ also. Assuming that we have done so we can see that the next application of LHR on $G(x)$ will give rise to the expression $$H(x) = \frac{1}{2}\left\{\frac{g(x)f''(x) - (1 + f'(x))g'(x)}{\{g(x)\}^{2}}\right\}\tag{4}$$ where $g(x) = (x + f(x))\cdot\log(x + f(x))\cdot\log\log(x + f(x))$.

Note that $g(x) \to e^{e + 1}$ so I hope that we don't need further application of LHR on $H(x)$ and the final limit would be $$\dfrac{A}{2e^{2e + 2}}\tag{5}$$ where $$A = \lim_{x \to 0}g(x)f''(x) - (1 + f'(x))g'(x)\tag{6}$$ We can now see that the real challenge is to evaluate $f'(x), f''(x), g'(x)$ and it would require a reasonable amount of calculation. I will have to leave my keyboard and go for a pen-paper calculation to handle this. Will post the calculation if I succeed.

The easiest part is $g'(x)$ given by $$\begin{aligned}g'(x) &= (1 + f'(x))\log(x + f(x))\log\log(x + f(x))\\ &\,\,\,\,+\,\, (1 + f'(x))\log\log(x + f(x))\\ &\,\,\,\,+\,\, (1 + f'(x))\\ &= ( 1 + f'(x))\{1 + \log\log(x + f(x)) + \log(x + f(x))\log\log(x + f(x))\}\end{aligned}$$ Based on the assumption made in bold we can see that $(1 + f'(x))g'(x)$ tends to $$(1 - e^{e + 1})^{2}(e + 2)\tag{7}$$ From $(1)$ we can see that $\log f(x) = p(x)\log p(x)$ where $p(x) = (1 + x)^{1/x}$. Hence \begin{align}\frac{f'(x)}{f(x)} &= p'(x)(1 + \log p(x))\notag\\ &= p(x)(1 + \log p(x))(\log p(x))'\notag\\ &= p(x)(1 + \log p(x))\left(\frac{\log (1 + x)}{x}\right)'\notag\\ &= p(x)(1 + \log p(x))\left(\frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)}\right)\tag{8}\end{align} Now we can see that $p(x) \to e$ so that $f'(x)/f(x)$ tends to $(2e)(-1/2) = -e$. Since $f(x) \to e^{e}$ we see that $f'(x) \to -e^{e + 1}$ and our assumption (mentioned in bold earlier) is verified (which at the same time verifies $(7)$).

Now the hardest part is to differentiate $(8)$ and obtain $f''(x)$. We have $$\begin{aligned}f''(x) &= f'(x)p(x)\{1 + \log p(x)\}\left(\frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)}\right)\\ &\,\,+\,\,f(x)\{1 + \log p(x)\}p(x)\left(\frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)}\right)^{2}\\ &\,\,+\,\,f(x)p(x)\left(\frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)}\right)^{2}\\ &\,\,+\,\,f(x)\{1 + \log p(x)\}p(x)\left(\frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)}\right)'\end{aligned}$$ Hence we can see that $f''(x)$ tends to $$e^{e + 2} + \frac{e^{e + 1}}{2} + \frac{e^{e + 1}}{4} + \frac{4e^{e + 1}}{3} = \frac{12e^{e + 2} + 25e^{e + 1}}{12}$$ The desired limit $A$ is thus $$\begin{aligned}A &= \frac{12e^{2e + 3} + 25e^{2e + 2}}{12} - (e + 2)(1 - e^{e + 1})^{2}\\ &= \frac{12e^{2e + 3} + 25e^{2e + 2} - 12(e + 2)(1 - 2e^{e + 1} + e^{2e + 2})}{12}\\ &= \frac{12e^{2e + 3} + 25e^{2e + 2} - 12(e - 2e^{e + 2} + e^{2e + 3} + 2 - 4e^{e + 1} + 2e^{2e + 2})}{12}\\ &= \frac{e^{2e + 2} + 24e^{e + 2} + 48e^{e + 1} - 12e - 24}{12}\end{aligned}$$ The final limit is $A/2e^{2e + 2}$ so that the final limit is given by $$\frac{e^{2e + 2} + 24e^{e + 2} + 48e^{e + 1} - 12e - 24}{24e^{2e + 2}} = \frac{1}{24} + \frac{2e^{e + 2} + 4e^{e + 1} - e - 2}{2e^{2e + 2}}$$ Note: The limit of $$a(x) = \frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)} = \dfrac{\dfrac{x}{1 + x} - \log(1 + x)}{x^{2}} = -\frac{1}{2} + \frac{2}{3}x + \cdots$$ and its derivative were calculated using the Taylor expression given above leading to $a(x) \to -1/2$ and $a'(x) \to 2/3$ as $x \to 0$.