Our plan of an answer is as follows. Firstly, we will introduce Planck's constant $\hbar$ so that the particular value $\hbar=1$ corresponds to the original problem. Secondly, we mention a connection to (what physicists often calls) the group property of Feynman path integrals. Thirdly, we will show that the sought-for formula happens to be the classical "instanton" contribution in a saddle point/steepest descent asymptotic expansion, which becomes valid as $\hbar\to 0$. We are currently unaware if semi-classical localization methods can be applied to justify the saddle point/steepest descent expansion, and we shall not attempt to make a justification here.
Now let us get to business. Define endpoints $x_0\equiv x>0$ and $x_{n+1}\equiv t>0$. We start by introducing Planck's constant $\hbar$ into the $u_n$ function in eq. (1.9) of arXiv:1102.4729,
$$u_n(x,t,\hbar)
~:=~\left[\prod_{j=1}^n 2 \int_0^{\infty} \frac{\mathrm{d}x_j}{\sqrt{\hbar}}\right]
\prod_{i=1}^{n+1}\frac{e^{-\frac{x_{i-1}^2}{2\hbar x_i}}}{\sqrt{2 \pi x_i}}.
\tag{1.9$\hbar$}
$$
In particular, for $n=0$, we have
$$ u_{n=0}(x,t,\hbar)
~=~\frac{e^{-\frac{x^2}{2\hbar t}}}{\sqrt{2 \pi t}}. $$
The $u_n$ function enjoys various scaling/homogeneity properties,
$$\begin{align} u_n( x,t,\hbar)
~=~&\sqrt{\lambda} u_n(\lambda x,\lambda t,\lambda\hbar) \cr
~=~& \lambda u_n(\lambda x,\lambda^{2^{n+1}} t,\hbar) \cr
~=~& \sqrt{\lambda} u_n( x,\lambda^{2^n} t,\frac{\hbar}{\lambda}),
\qquad \lambda~>~0. \end{align} \tag{H} $$
With the help of the first homogeneity property in eq. ($H$), we can immediately deduce the corresponding $\hbar$ generalization of eq. (3.14) in arXiv:1102.4729,
$$ \lim_{n\to \infty}u_n(x,t,\hbar)
~=~\frac{1}{\sqrt{\hbar}}e^{-\frac{2x}{\hbar}}. \tag{3.14$\hbar$} $$
So the question is basically how do we derive, understand, motivate, etc., eq. (3.14$\hbar$) physically? To get to a path integral interpretation, we note that the $u_n$ function has (what physicists often call) a group property,
$$ u_{n+1+m}(x,z,\hbar) ~=~ 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{n}(x,y,\hbar)u_{m}(y,z,\hbar), \tag{G}
$$
in close analogy with the Feynman propagator $K(x_f,t_f;x_i,t_i)$ with
$$ K(x_3,t_3;x_1,t_1)
~=~ \int_{-\infty}^{\infty}\!\mathrm{d}x_2 ~ K(x_3,t_3;x_2,t_2) K(x_2,t_2;x_1,t_1).$$
So the "sum of histories" from $x$ to $z$ can be calculated by integrating over an intermediate point $y$. The $n$ in the $u_n$ function plays the role of a discretized time variable. As a consistency check, it is easy to see (by performing some elementary integrals) that the right-hand side of eq. (3.14$\hbar$),
$$u_{n=\infty}(x,t,\hbar)
~=~\frac{1}{\sqrt{\hbar}}e^{-\frac{2x}{\hbar}} \qquad
\left(~\to~ \sqrt{2\pi x} \delta(x)
\quad \mathrm{for} \quad
\hbar ~\to~ 0\right), $$
does indeed solve the group equation $(G)$ in the particular cases $n,m=0,\infty$,
$$\begin{align} u_{\infty}(x,z,\hbar) ~=~& 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{\infty}(x,y,\hbar) u_{\infty}(y,z,\hbar) \cr
~=~& 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{\infty}(x,y,\hbar) u_{0}(y,z,\hbar)\cr
~=~& 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{0}(x,y,\hbar) u_{\infty}(y,z,\hbar).\end{align}
$$
Next introduce Gaussian momenta $p_1, \ldots,p_{n+1},$ with
$$\int_{-\infty}^{\infty}\frac{\mathrm{d}p_i}{2\pi\sqrt{\hbar}}e^{-\frac{1}{2\hbar}x_ip_i^2}
~=~ \frac{1}{\sqrt{2 \pi x_i}}, \qquad x_i~>~0.$$
Then the $u_n$ function becomes
$$u_n(x,t,\hbar) ~=~ \left[\prod_{j=1}^n 2\int_0^{\infty}\frac{\mathrm{d}x_j}{\sqrt{\hbar}}\right]\left[\prod_{i=1}^{n+1}\int_{-\infty}^{\infty}\frac{\mathrm{d}p_i}{2\pi\sqrt{\hbar}}\right]e^{-\frac{S}{\hbar}},
$$
with Euclidean phase space action
$$S~:=~\frac{1}{2}\sum_{i=1}^{n+1}\left(\frac{x_{i-1}^2}{x_i}+x_i p_i^2\right).$$
Now let us turn to the saddle point/steepest descent asymptotic expansion.
The classical equations of motion are
$$ 0 ~\approx~ \frac{\partial S}{\partial p_i}
~=~ x_i p_i, $$
$$ 0 ~\approx~ \frac{\partial S}{\partial x_i}
~=~ \frac{x_i}{x_{i+1}}-\frac{x_{i-1}^2}{2x_i^2} + \frac{p_i^2}{2}, $$
where we use $\approx$ sign instead of $=$ sign to emphasize when classical equations of motion have been applied. The classical solution is
$$p_i ~\approx~ 0, \qquad q_i ~\approx~ q_{i-1}^2,$$
where we have defined $q_i :=\frac{x_i}{2x_{i+1}}$.
So $q_i \approx q_{i-1}^2 \approx q_{i-2}^4 \approx \ldots \approx q_0^{2^i}$. Now the telescopic product
$$\prod_{i=0}^n 2q_i ~=~\prod_{i=0}^n\frac{x_i}{x_{i+1}}~=~\frac{x_0}{x_{n+1}}=\frac{x}{t},$$
is fixed by boundary conditions $x$ and $t$. So
$$q_0^{2^{n+1}-1}~=~q_0^{\sum_{i=0}^n 2^{i}}
~\approx~\prod_{i=0}^n q_i ~=~\frac{x}{2^{n+1}t},$$
and therefore the unique classical solution is
$$q_i ~\approx~ \left( \frac{x}{2^{n+1}t} \right)^{\frac{2^i}{2^{n+1}-1}} ~\to~ 1 \qquad
\mathrm{for} \qquad n ~\to ~\infty. $$
Hence classically $x_i \approx 2^{-i}x$ for $n=\infty$. The classical value of the action is
$$ S_{\mathrm{cl}} ~\approx~ \sum_{i=0}^{n} x_i q_i ~\to~ x\sum_{i=0}^{\infty}2^{-i} ~=~2x \qquad
\mathrm{for} \qquad n ~\to~ \infty, $$
so the classical "instanton" contribution $e^{-\frac{S_{\mathrm{cl}}}{\hbar}}$ happens to be the right-hand side of eq. (3.14$\hbar$), up to a $\sqrt{\hbar}$ factor. This is our main observation.
A more complete treatment would now calculate the one-loop Van Vleck determinant $\det(\partial^2S)$ in the saddle point/steepest descent asymptotic expansion. Here we will only make a couple of further remarks. The Hessian $\partial^2 S$ of the action is
$$ \frac{\partial^2 S}{\partial x_i\partial x_j}
~=~\delta_{i,j}\left(\frac{1}{x_{i+1}}+\frac{x_i^2}{x_{i+1}^3}\right)-\delta_{i+1,j}\frac{x_i}{x_j^2}-\delta_{i-1,j}\frac{x_j}{x_i^2}, $$
$$ \frac{\partial^2 S}{\partial p_i\partial x_j}
~=~ \frac{\partial^2 S}{\partial x_i\partial p_j} ~=~ \delta_{i,j}p_i\approx 0, \qquad
\frac{\partial^2 S}{\partial p_i\partial p_j} ~=~ \delta_{i,j}x_i. $$
Since we have $n+1$ momenta $p_i$, but only $n$ positions $x_i$, we would naively expect the Van Vleck determinant $\det(\partial^2S) \sim x$ to be proportional to $x$ on-shell. This would mean a $1/\sqrt{x}$ factor in the expansion. It would be interesting to see a detailed calculation of the Van Vleck determinant $\det(\partial^2S)$.
