When you open the tube, yes, air will be drawn into it. This is indirectly due to gravity, in the sense that what will drive the flow is a pressure gradient. The pressure in the tube is lower than the pressure outside (after you pump air out and let things settle down, this is true no matter where you open the hole), so air is drawn in. Gravity has a role in maintaining the pressure gradient, as air is drawn into the tube, it is replaced and the pressure outside remains $\sim$ constant because there is much more air outside the tube than in, and gravity maintains a pressure gradient in the atmosphere.
If you want a rough idea of how long this will take, the rule of thumb for the propagation of pressure waves (i.e. disturbances to an equilibrium state) is that they move at the sound speed of the fluid. The characteristic timescale for your tube will therefore be set by the sound speed inside it and its length. Keep in mind that the sound speed inside will vary along the length, and as a function of time as the density increases, so getting an accurate number will be difficult, but you can estimate upper and lower bounds by solving for the sound speed along the tube before you open the hole and after the tube equilibrates with the outside (this will involve some integrals).
If you want a full blown solution (e.g. to run a simulation), you'll be solving the Navier-Stokes equation:
$$\rho\frac{D\mathbf{v}}{Dt} = -\nabla p + \nabla\cdot\mathbf{T}+\mathbf{f}$$
You'd probably set this up as a tube in a simulation box where the box boundaries are fixed to a model of the atmospheric pressure, density, etc. gradients. You could also model just the tube to simplify away how air flows into it from the region around the hole; in this case your boundary condition is simply to fix the pressure, density, etc. to the external values at the altitude of the hole. This is probably a good starting point to try in 1D.
Be sure to find a good book or other reference on fluid dynamics that covers this equation before attempting to solve it as it can be a bit tricky - for instance the general case cannot be solved without resorting to numerical approximations. Also make sure you're familiar with the notation, e.g. the $\frac{D}{Dt}$ is a material derivative.
