We have to make a few assumptions here. The first would be that the soap solution mixes with the water evenly and instantaneously in the capillary - this would be a good approximation, as it's a fairly small volume in the tube, on a similar order of magnitude to the volume of the drop. The next is the form of the meniscus: Please excuse my crude diagram! You can see that the meniscus is approximated to have a flat bottom, with the sloped sides making an angle $\theta$ - this is symmetrical. 707
The difference in the liquid height from the surface of the body to the meniscus will be $h$. The capillary will have a radius of $r$.
We can take the difference in the surface free energy by changing the height as
$2\pi rh(\gamma_{sl}-\gamma_{sg})$;
$\gamma_{sl}$ is the surface free energy of the solid-liquid interface, and $\gamma_{sg}$ is the surface free energy of the solid-gas interface.
Then, we also take the difference in the gravitational potential energy in changing the height of the liquid:
$\pi r^2h\rho\times\frac{h}{2}g$,
where $\rho$ is the density of the liquid, and we take the centre of mass of the liquid in the capillary to be at $h/2$.
So, the free energy of the meniscus is given by
$\frac{\pi}{2}r^2h^2\rho g+2\pi rh(\gamma_{sl}-\gamma_{sg})$.
We have a stable meniscus when this is minimised, so we differentiate it with respect to $h$, and set that equal to zero, giving
$h=\frac{-2(\gamma_{sl}-\gamma_{sg})}{\rho gr}$.
We can then substitute in the Young Equation, finally giving
$h=\frac{2\gamma_{lg}cos\theta}{\rho gr}$.
The surface free energy of the liquid-gas interface, $\gamma_{lg}$, will be determined by the soap, which would also alter $\rho$. Also, note that $\theta$ can be negative - in this case the level in the capillary would fall, which would be the case with soap using the numbers you gave.
With these assumptions, the water level would change near instantaneously - though it may require some mixing. 707