It looks like you're trying to use area to determine how much back-flow there will be. The flow will be depended on the pressure gradients, which are dependent not just on cross sectional area but also length.
It sounds like you're trying to model something like a 3D printer filament being pushed through a heated nozzle and want to determine how tight of a tolerance you'll need to prevent melted plastic from flowing back out the entrance.
This is a pretty complicated problem but I'll try to give the basics of it:
First we need to establish some pressures, and additional dimensions:
Now we can see the melted piston in green can flow in two directions back around the solid piston, or forward through the exit orifice. First let's look at the forward flow rate:
For fairly viscous media (like melted plastic) through thin pipes flow rate is calculated by the Hagen–Poiseuille equation:
$$Q= \frac{\pi\, R^4\,\Delta P}{8\,\mu\, L}$$
Where $Q$ is your volumetric flow rate, $L$ is the length of your tube ($L_0$), $R$ is your tube radius ($r_0$), $\mu$ is your viscosity (going to be highly temperature dependent), and $\Delta P$ is the difference in pressure across the tube ($P_1-P_2$)
So plugging in:
$$Q_0= \frac{\pi\, r_0^4\,(P_1-P_0)}{8\,\mu\, L_0}$$
Flow around a piston can be modeled like flow between two parallel plates. Except the plates are wrapped into a circle to form cylinders. When one plate is moving relative to the other this is called Couette Flow.
$$Q=\frac{\Delta P\,h^3 \, w }{12 \mu\,L} - \frac{U\,h\,w}2$$
Where $h$ is the height of our chanel $r-r_1$, $w$ is the width of our channel (in this case circumference of the piston) $2\pi r_1$, $U$ is the velocity of our piston, $\Delta P$ is $P_1-P_2$, and $L$ is $L_1$
Plugging in we get:
$$Q_1=\frac{\pi(P_1-P_2)\,(r-r_1)^3 \, r_1 }{6 \mu\,L_1} - \pi\,U\,(r-r_1)\,r_1$$
Now we're looking for this second Q to be zero. This means all of the melted material gets pushed out the front hole. If the flow rate is positive, the edge of the melted plastic would flow further up the channel, increasing $L_1$ which would in turn decrease the flow rate. This is good news, it means as long as the sleeve is long enough the flow will stabilize around zero flow. Also since we are now assuming all the newly melted plastic is going to go out the exit, this allows us to relate the flow rate out the front hole with the velocity of the piston:
$$Q_0=U\, \pi {r_1}^2$$
Combining the equations:
$$0=Q_1=\frac{\pi(P_1-P_2)\,(r-r_1)^3 \, r_1 }{6 \mu\,L_1} - \frac{Q_0\,(r-r_1)}{r_1}$$
$$0=\frac{\pi(P_1-P_2)\,(r-r_1)^3 \, r_1 }{6 \mu\,L_1} - \frac{\pi\, r_0^4\,(P_1-P_0)\,(r-r_1)}{8\,\mu\,r_1 \, L_0}$$
Now, while viscosity will probably be different in the different regions due to temperature, it's super convenient if we ignore that because then we can just multiply it out.
$$\frac{(P_1-P_2)\,(r-r_1)^3 \, r_1 }{3\,L_1} = \frac{r_0^4\,(P_1-P_0)\,(r-r_1)}{4\,r_1 \, L_0}$$
Further rearranging:
$$4(P_1-P_2)\,(r-r_1)^2 \, {r_1}^2 \,L_0 = 3\,{r_0}^4\,(P_1-P_0)\,L_1$$
Similarly if ($P_1=P_2$) we can cancel that out:
$$4\,(r-r_1)^2 \, {r_1}^2 \,L_0 = 3\,{r_0}^4\,L_1$$
$$r=r_1+\frac{{r_0}^2}{r_1}\sqrt{\frac{3\,L_1}{4\,L_0}}$$
This will give you the maximum size bore you can use with the given lengths and other sizes of your hole.
