I was thinking about this and I think I have an answer for you:
Consider the probability of getting $n_{for}$ votes out of $n$ observations with our base assumption of independence. For a given $p$ the probability of such an observation is:
$P(p)=$$\frac{\Gamma (n+2)}{\Gamma (n_{for}+1) \Gamma (n-n_{for}+1)}$$p^{n_{for}}(1-p)^{n-n_{for}}$
We then proceed to calculate the probability by summing over all possible states. As you guessed there is an edge case where we have $n_{for}>N/2$ or $n-n_{for}>N/2$ where our vote is already decided. If we aren't in such a case we can evaluate the sum/integral:
$\sum_{i=Ceiling(N/2)-n_{for}} ^{N-n} \int_0 ^1 dp (P(p) \binom{N-n}{i} (p)^i (1-p)^{N-n-i})$
Evaluating the integral yields:
$\sum_{i=Ceiling(N/2)-n_{for}} ^{N-n} \frac{\Gamma (n+2) \Gamma (i+n_{for}+1) \Gamma (N-n+1) \Gamma (-i+N-n_{for}+1)}{\Gamma (i+1) \Gamma (N+2) \Gamma (n_{for}+1) \Gamma (n-n_{for}+1) \Gamma (-i+N-n+1)}$
And finally using a computer I was able to get the following expression:
$\frac{\pi \Gamma (n+2) \Gamma \left(\left\lceil \frac{N}{2}\right\rceil +1\right) \Gamma (N-n+1) \, _3\tilde{F}_2\left(1,\left\lceil \frac{N}{2}\right\rceil +1,-N+n-n_{for}+\left\lceil \frac{N}{2}\right\rceil ;\left\lceil \frac{N}{2}\right\rceil -N,-n_{for}+\left\lceil \frac{N}{2}\right\rceil +1;1\right)}{\Gamma (N+2) \Gamma (n_{for}+1) \Gamma (n-n_{for}+1) \Gamma \left(N-n+n_{for}-\left\lceil \frac{N}{2}\right\rceil +1\right)}$
where $\tilde{F}$ is the regularized generalized hypergeometric function.
This would be the probability of not switching so subtract the above from 1 to get the probability of switching.
If we know p ahead of time then this is actually a trivial problem:
$\sum_{i=ceiling{N/2}-n_{for}} ^{N-n} p^i (1-p)^{N-n-i} \binom{N-n}{i} =$
$ p^{\left\lceil \frac{N}{2}\right\rceil - n_{for}} \binom{N-n}{\left\lceil \frac{N}{2}\right\rceil -n_{for}} (1-p)^{-\left\lceil \frac{N}{2}\right\rceil +N-n+n_{for}} \, _2F_1\left(1,-N+n-n_{for}+\left\lceil \frac{N}{2}\right\rceil ;-n_{for}+\left\lceil \frac{N}{2}\right\rceil +1;\frac{p}{p-1}\right)$