Evaluate boolean circuit on batch of similar inputs
I'd consider it unlikely that such a trick is easy to find and/or will give you significant gains, as it would give nontrivial satisfiability algorithms. Here's how:

First of all, while ostensibly easier, your problem can actually solve the more general problem of, given a circuit $C$ and $N$ inputs $x_0, \ldots, x_{N-1}$, evaluate $C$ at all the inputs faster than $\tilde{O}(N\cdot|C|)$ time. The reason is that we can tweak $C$ into a circuit $C'$ of size $|C| + \tilde{O}(Nn)$ which, on input $0^i10^{N-1-i}$, outputs $C(x_{i})$. Basically, we just make a little lookup table that sends $0^i10^{N-1-i}$ to $x_i$, and wire it into $C$.

Nontrivial algorithms for batch evaluation of boolean-circuits can then be used to make fast satisfiability algorithms. Here's an example in the simple case where we suppose we have an algorithm doing evaluation in time $\tilde{O}(|C|^{2-\epsilon} + (N\cdot|C|)^{1-\epsilon/2} + N^{2-\epsilon})$ for any constant $\epsilon &gt; 0$. On input a circuit $C$, we can decide satisfiability by expanding $C$ into a circuit $C'$ of size $2^{n/2}\cdot |C|$ which is just the OR over all possible choices of the first $n/2$ inputs to $C$ (leaving the other inputs free). We then batch-evaluate $C'$ on all of its $2^{n/2}$ inputs. The end result is that we find a satisfying assignment to $C'$ iff $C$ is satisfiable. The running time is $\tilde{O}(2^{(n/2)(2-\epsilon)}\cdot |C|^{2-\epsilon}) = \tilde{O}(2^{n(1-\epsilon/2)}\cdot \textrm{poly}(|C|))$.