I have a piston of diameter 0.7 mm exactly and a 0.7 mm drill bit. Let's call the diameter of the piston $r\_1$, the diameter of a hole at the end of a tube $r\_0$, and the final-actual diameter of the tube hole $r$. Then with fluid being pushed by the piston through the hole at the end ($r\_0$) I calculate that I need:

$$
\pi (r^2 - r\_1^2) \lt \pi r\_0^2 \\
\iff r^2 \lt r\_1^2 + r\_0^2 \\
\iff r \lt \sqrt{r\_1^2 + r\_0^2}
$$
in order to prevent back-flow in the negative direction of the piston movement. I'm ignoring gravity. With $r\_0 = $0.05mm, $r\_1 = $0.35mm, I come up with:
$r \lt \sqrt{(0.05)^2 + (0.35)^2} \approx 0.3536$. So can I get a hole that's between 0.7 and 0.7072 mm (preferably closer to 0.7 mm) using a 0.7 mm bit?

It's okay if the tolerance is so tight that the piston gets stuck at room temp, because the piston is a rod of continually fed material that will be melted in the tube.