By analogous reasoning to proving the formula for the pmf of a Binomial random variable, we get $$\sum_{\substack{S\subseteq[100]\\|S|=65}}\prod_{i\in S}p_{i}\prod_{i\in[100]\setminus S}(1-p_{i}),$$ where the sum is over all subsets $S$ of $[100]=\{1,2,\ldots,100\}$ with size $65.$ It's not terribly clear how useful this formula will be, but if you had some bounds on the $p_{i},$ you might be able to use this to prove that the probability lies in some interval.
For instance, if all $p_{i}\in[1/2,3/4],$ then $\prod_{i\in S}p_{i}\in[(1/2)^{65},(3/4)^{65}],$ and $\prod_{i\in[100]\setminus S}(1-p_{i})\in[(1/4)^{35},(1/2)^{35}]$, giving a product in $[(1/2)^{65}(1/4)^{35},(3/4)^{65}(1/2)^{35}]$ for all $S\subseteq[100].$ Then the probability lies in the interval $[\binom{100}{65}(1/2)^{135},\binom{100}{65}(3/4)^{65}(1/2)^{35}]\approx[2.514\times 10^{-14},1]$, which admittedly is not super helpful, but with tighter bounds, this could be improved.