This model is called the Wright-Fisher model for genetic drift (when $n$ is even). Each gene in a population is a copy of a random gene from the previous generation. You are looking at the case that a mutation occurs at time $0$, and then you are asking how long it takes for the gene to be fixed. The following argument seems to be standard:
Let $D_t$ be the count of pairs of genes of different types. $D_t = b_t(n-b_t)$. So, this starts at $n-1$, it is always at most $n^2/4$, and it is $0$ when the gene is fixed.
$$\begin{eqnarray}E[D_{t+1}|b_t] &=& {n \choose 2}\times 2\frac{b_t}{n}\frac{n-b_t}{n} \newline &=& \frac{n^2-n}{n^2} b_t(n-b_t) \newline &=&(1-\frac{1}{n})D_t\end{eqnarray}$$
Since $D_0 = n-1, E[D_t] = (n-1)(1-\frac{1}{n})^t.$
We can use this for easy upper and lower bounds on the probability that the population has been fixed. or equivalently that it has not been fixed. Let $\tau$ be the first time the population is fixed. If $\tau \le t$ then $D_t=0$. If $\tau \gt t$ then $n-1 \le D_t \le n^2/4$.
$$\begin{eqnarray}P(\tau \gt t) (n-1) &\le & ~~~~E[D_t] &\le & P(\tau \gt t) \frac{n^2}{4} \newline \bigg(1-\frac{1}{n}\bigg)^{-t}&\le & P(\tau > t)^{-1} &\le & \frac{n^2}{4(n-1)} \bigg(1-\frac{1}{n}\bigg)^{-t} \newline \bigg(1-\frac{1}{n}\bigg)^t& \ge& P(\tau > t) &\ge & \frac{4(n-1)}{n^2} \bigg(1-\frac{1}{n}\bigg)^t.\end{eqnarray}$$
You ask about the probability that the gene is fixed in one direction. The probability that the mutation takes over the population is $1/n$, so a lower bound for the probability that $b_t = 0$ is $1-(1-\frac{1}{n})^t-1/n$.
While the estimate
$\frac{D_0}{n-1}\bigg(1-\frac{1}{n}\bigg)^t \ge P(\tau \gt t) \ge \frac{4D_0}{n^2} \bigg(1-\frac{1}{n}\bigg)^t$
is reasonable when $b_0 = \Theta(n)$, it's not great if you fix $b_0=1$ as $n\to \infty$. $\bigg( 1-\frac{1}{n} \bigg) ^t \approx e^{-t/n}$. This estimate gets weaker, but the chance that $b_t=0$ increases with $n$. In the limit $n\to \infty$ we can consider the map $b_{t+1} = \textrm{Pois}(b_t)$. As $t\to \infty, P(b_t=0) \to 1$. This can be computed exactly:
$$\begin{eqnarray}P(b_1 = 0) &=& e^{-1} \newline P(b_2 = 0) &=& e^{e^{-1}-1} \newline P(b_3 = 0) &=& e^{e^{e^{-1}-1}-1} \newline P(b_{t+1} = 0) &=& \sum_{k=0}^\infty \frac{1}{e~ k!}P(b_t=0)^k = \exp(P(b_t=0)-1)\end{eqnarray}$$
$e^{x-1}$ has a parabolic fixed point at $1$, with expansion $1 + (x-1) + \frac{1}{2}(x-1)^2 + o((x-1)^2)$. This indicates that the convergence to $1$ is something like $1-c/t$. Since $P(b_{1000} = 0) = 0.998008$ it appears that $c = 2$.
