Estimation of upper bound.
L signs the value of the expression, then we can write:
\begin{align}L^2=1+\left(1+(1+\cdots+(1+(1+s)^{{1/n}})^{{1/{n-1}}}\cdots)^{1/4}\right)^{1/3}\end{align}
where
\begin{align}s=\sqrt[n+1]{1+\sqrt[n+2]{1+ \sqrt[n+3]{1+ \cdots}}} \tag{1}\end{align}
Accordance with Bernoulli$^,$s inequality if $0\le{r}\le1$
and $x\ge-1$ then $(1+x)^r\le(1+xr)$
In our case $r=\frac{1}{k}, k=3,4, ....n $ and $s\gt1 $.
First apply to $(1+s)^{1/n}$ we have that $\le1+\frac{s}{n}$
then outward apply to all power factor we get the following:
\begin{align}L^2\le1+1+\left(1+(1+\cdots+(1+(1+\frac{s}{n}))\frac{1}{n-1}\cdots)\frac{1}{4}\right)\frac{1}{3}\end{align}
Perform the multiplications:
\begin{align}L^2\le2 +2(\frac{1}{3!} +\frac{1}{4!}+......\frac{1}{(n-1)!}+\frac{s}{n!})\end{align}
Namely
\begin{align}L^2\le2 +2\sum_{k=3}^n\frac{1}{k!}\end{align}
We could apply to $s$ the above procedure then we have:
\begin{align}L^2\le2 +2\sum_{k=0}^{\infty}\frac{1}{k!}-5\end{align}
\begin{align}L\le \sqrt{2e-3}=1,56094\end{align}