Let $Y_0 \sim N(0,1)$. Then $Y_1 = \left|Y_0\right|$ follows the half-normal distribution and $\overline{Y_1} = \sqrt{\frac {2}{\pi}}$. This is considered one iteration of a "flipping". A second iteration of a flipping would be $Y_2 = \left|Y_1 - \sqrt{\frac {2}{\pi}}\right| + \sqrt{\frac {2}{\pi}}$. Thus, the $i$th flipping can be defined as: $$Y_i = \left|Y_{i-1} - \overline{Y_{i-1}}\right| + \overline{Y_{i-1}}\text{ ; } Y_0 \sim N(0,1)$$
What is $\overline{Y_i}$ as $i \to \infty$? Also, can $\overline{Y_i}$ be expressed as an explicit function?
I believe that $\overline{Y_i} = O\left(\ln{i}\right)$ so $\displaystyle{\lim_{i \to \infty}} \overline{Y_i} = \infty$ based on the following logarithmic function I fitted after a simulation of $10000$ "flippings". But, I don't know how to prove this. Note that $\overline{Y_i}$ eventually grows slower than the logarithmic function in the following plot, so it may converge also.
The details of this are probably intractable, but the long-term behaviour can be analyzed. After many reflections, almost all the probability is concentrated near the expected value, and only an improbable unreflected tail remains. We can model this as a delta distribution at $y$ plus the tail at $x\gt y$. The expectation of this distribution is
$$ y+\int_y^\infty(x-y)\frac{\mathrm e^{-\frac{x^2}2}}{\sqrt{2\pi}}\mathrm dx=y+\frac1{\sqrt{2\pi}}\left(\mathrm e^{-\frac{y^2}2}-y\int_y^\infty\mathrm e^{-\frac{x^2}2}\mathrm dx\right)\;. $$
With
\begin{eqnarray*} \int_y^\infty\mathrm e^{-\frac{x^2}2}\mathrm dx &=& \int_y^\infty\frac1x\cdot x\mathrm e^{-\frac{x^2}2}\mathrm dx \\ &=& \left[-\frac1x\mathrm e^{-\frac{x^2}2}\right]_y^\infty-\int_y^\infty\frac1{x^2}\mathrm e^{-\frac{x^2}2}\mathrm dx \\ &=& \frac1y\mathrm e^{-\frac{y^2}2}-\int_y^\infty\frac1{x^3}\cdot x\mathrm e^{-\frac{x^2}2}\mathrm dx \\ &=& \frac1y\mathrm e^{-\frac{y^2}2}+\left[\frac1{x^3}\mathrm e^{-\frac{x^2}2}\right]_y^\infty+3\int_y^\infty\frac1{x^4}\mathrm e^{-\frac{x^2}2}\mathrm dx \\ &=& \frac1y\mathrm e^{-\frac{y^2}2}-\frac1{y^3}\mathrm e^{-\frac{y^2}2}+O\left(\frac1{y^5}\mathrm e^{-\frac{y^2}2}\right)\;, \end{eqnarray*}
this becomes
$$ y+\frac1{\sqrt{2\pi}}\frac1{y^2}\mathrm e^{-\frac{y^2}2}+O\left(\frac1{y^4}\mathrm e^{-\frac{y^2}2}\right)\;. $$
Reflecting the delta distribution in this expectation shifts it forward by approximately $\sqrt{\frac2\pi}\frac1{y^2}\mathrm e^{-\frac{y^2}2}$. If we consider $y$ as a function of the step count $t$, this yields the approximate differential equation
$$ \dot y=\sqrt{\frac2\pi}\frac1{y^2}\mathrm e^{-\frac{y^2}2}\;, $$
and then separating variables leads to
$$ \int y^2\mathrm e^{\frac{y^2}2}\mathrm dy=\sqrt{\frac2\pi}\int\mathrm dt\;. $$
This could be analyzed in more detail, but if we take logarithms, the dominant contributions will be $\frac{y^2}2$ on the left and $\log t$ on the right, so we get
$$ \frac{y^2}2\approx\log t $$
and thus
$$ y\approx\sqrt{2\log t}\;, $$
which confirms your observation of sublogarithmic behaviour.
Since you tried to improve on the approximation in a comment (presumably to achieve better agreement with your data), here are three improvements.
On the right, there’s an additional term $\log\sqrt{\frac2\pi}$, and $t$ is really $t-t_0$ with a free integration constant $t_0$. On the left, we have
$$ \int y^2\mathrm e^{\frac{y^2}2}\mathrm dy=y\mathrm e^{\frac{y^2}2}+O\left(\frac1y\mathrm e^{\frac{y^2}2}\right)\;. $$
Solving the resulting equation
$$ 2\log y+y^2=\log\frac2\pi+2\log(t-t_0) $$
yields
$$ y=\sqrt{\log\frac2\pi+2\log(t-t_0)} $$
in a first approximation, and substituting that into $\log y$ yields an improved approximation
$$ y=\sqrt{\log\frac2\pi+2\log(t-t_0)-\log\left(\log\frac2\pi+2\log(t-t_0)\right)}\;. $$
Even without fitting $t_0$, setting $t_0=0$ again, this is closer to your data, e.g. about $3.88$ instead of $4.24$ at $t=10000$.