The simplest model for the vibrations in a damped string is the damped wave equation:
$$ \frac{\partial^2y}{\partial t^2}y +2b\frac{\partial y}{\partial t}=c^2\frac{\partial^2y}{\partial x^2} $$
where $y$ is the wave amplitude, $t$ is time, $x$ is the position coordinate along the string, $c$ is the speed of the waves in the string (which depends on the string material and the tension) and $2b$ is a damping consant. You don't need to understand exactly what this equation means and how to solve it, but it models the damping force acting on a unit length of each piece of the string as being proportional to its speed, and $2b$ is the proportionality constant. For a string of length $L$ fixed at the end points $x = 0$ and $x = L$, and a sufficiently small $b$, the standing wave solutions are
$$y_m(x,t) = e^{-bt} \cos\left(\sqrt{\omega_m^2 - b^2}t+\phi_m\right)\sin\left(\frac{\omega_m}{c}x\right) $$
where $m$ numbers the different modes ($m=1$ being the fundamental), $\phi_m$ is some constant phase and
$$\omega_m = m\frac{c\pi}{L}. $$
$e^{-bt}$ is the factor governing how the amplitude decays over time, and the decay is exponential as you guessed. However, in this simple model the time constant of the decay is $b^{-1}$ (the half-life is $b^{-1}\ln2$) which is the same for all modes. This may be more or less true for an experiment such as yours, but especially at very high frequencies I suspect this may no longer be the case.
I would expect in most strings the damping force may have a stronger dependence on velocity than what this simple model assumes (such as an additional velocity squared dependence due to air drag), resulting in a greater damping force and a faster decay in amplitude at high frequencies. Perhaps someone will chime in with a more sophisticated model for a damped string, or with experimental results.
