This is nothing more than a long comment, but I found myself wondering if there was an easy way to get a reasonable upper bound on the value of $Q$, assuming its limit exists. (Lower bounds are a dime a dozen; any truncation of the nested radical will do.) Here's what I came up with:
$$\begin{align}
\sqrt2Q&=
\sqrt2\sqrt{1!+\sqrt{2!+\sqrt{3!+\cdots}}}\\
&=\sqrt{2+2\sqrt{3!+\sqrt{4!+\cdots}}}\\
&=\sqrt{2+\sqrt{4\cdot2!+4\sqrt{3!+\cdots}}}\\
&=\sqrt{2+\sqrt{4\cdot2!+\sqrt{16\cdot3!+\sqrt{256\cdot4!+\cdots}}}}\\
&\gt\sqrt{2+\sqrt{3!+\sqrt{4!+\sqrt{5!+\cdots}}}}\\
&=Q^2-1
\end{align}$$
so $Q^2-\sqrt2Q-1\lt0$, which implies
$$Q\lt{\sqrt2+\sqrt6\over2}\approx1.93$$
This bound, while crude because crudely obtained, is nonetheless not too far from the reported value, $Q\approx1.827$.
Added later: It seems worth giving a proof that the sequence $Q_n=\sqrt{1!+\sqrt{2!+\sqrt{3!+\cdots+\sqrt n!}}}$ converges.
It's clear that the sequence is monotonically increasing, so it suffices to show it's bounded above. The proof is by induction (on $n$) of the following statement: For all $m,n\in\mathbb{N}$,
$$\sqrt{m!+\sqrt{(m+1)!+\cdots+\sqrt{(m+n)!}}}\le m!+1$$
The inequality is certainly for all $m$ in the base case $n=0$: $\sqrt{m!}\le m!+1$. Induction now says that
$$\sqrt{m!+\sqrt{(m+1)!+\cdots+\sqrt{(m+n)!}}}\le\sqrt{m!+((m+1)!+1)}$$
so it's enough to check that
$$m!+(m+1)!+1\le(m!+1)^2$$
which is easy enough to see, since
$$(m!+1)^2=m!m!+2m!+1=m!+(m!+1)m!+1\ge m!+(m+1)m!+1=m!+(m+1)!+1$$
Letting $m=1$ in the inequality $\sqrt{m!+\sqrt{(m+1)!+\cdots+\sqrt{(m+n)!}}}\le m!+1$, follows that $Q_n\le1!+1=2$ for all $n$, so the (monotonically increasing) sequence is bounded above, hence converges to a limit.