Consider the following thought experiment that will aid your understanding of the problem. Suppose you have a horizontal capillary with radius r_0 connected to a reservoir of liquid. Let the liquid level in the reservoir be at a height h above the capillary tube. At some instant in time let the gas-liquid interface in the capillary be at a length L from the reservoir. Now assuming laminar flow in the capillary tube and neglecting edge effects then the volumetric flow rate in the tube will be Q=\pi \Delta P/(8\mu L), where \Delta P=P-P_L. Here P_L is the pressure in the liquid just inside the liquid meniscus which has a contact angle \theta with the capillary tube wall, and P is the pressure at the inlet of the capillary.
At the gas-liquid interface the pressure P_L is related to the capillary pressure due to surface tension:
P_0-P_L=2 \sigma \cos(\theta)/r_0
where P_0 is the pressure in the ambient gas phase.
Recall that P is the pressure at the entrance of the capillary. This pressure is related to the hydrostatic pressure in the tank ( again neglecting edge effects).
P=\rho g h-P_0
where P_0 is the pressure in the ambient gas phase. Using the above expressions one can find the flow rate in the capillary:
Q=\frac{\pi}{\8 \mu L}(\rho g h +2\sigma \cos(\theta)/r_0)
Note that depending on the contact angle (wetting vs non wetting) it is possible to have zero flow rate in the capillary.
From the conservation of mass one can relate L to h by integrating the following
-A \rho dh/dt=\pi r_0^2 \rho dL/dt
One can then find when the liquid in the tube start flowing by increasing the liquid level in the reservoir. Again it does depend on the contact angle wetting versus non wetting.