Variation of fixation probability over time in Wright-Fisher model
The function probably look like $r(t) = A + B e^{-C t}$, where $r(t)$ is the rate of fixation at time $t$ (in generation). The reason for this exponential is that the probability that a mutation happens after $t$ generation is given by the exponential distribution.

As the rate of fixation at equilibrium must be at $2N\mu \frac{1}{2N} = \mu$, $A=\mu$, that is $r(t) = \mu + B e^{-C t}$

As at $t=0$, the rate must be 0. As $e^{0} = 1$, you should have $B=-\mu$. Hence, 

$$r(t) = \mu - \mu e^{-C t}$$

$C$ here represents the steepness of the curve. I don't know how you could calculate it a priori but it must depends upon your population size $N$. As the expected time to fixation starting at frequency $p_0$ is (from Kimura and Ohta 1968)

$$\bar t(p_0)=-4N\left(\frac{1-p_0}{p_0}\right)\ln(1-p_0)$$

For $p_0 = \frac{1}{2N}$

$$\bar t\left(\frac{1}{2N}\right)=-4N\left(2N-1\right)\ln\left(1-\frac{1}{2N}\right)$$

or

$$\bar t\left(\frac{1}{2N}\right)=\left(4N-8N^2\right)\ln\left(\frac{2N-1}{2N}\right)$$

if you prefer. Maybe $C = \bar t\left(\frac{1}{2N}\right)$ leading to

$$r(t) = \mu \left(1 - e^{-\left( \left(4N-8N^2\right)\ln\left(\frac{2N-1}{2N}\right) \right) t}\right)$$

Let me know if it matches (I'd pretty amazed)!