Assign few binary variables to make all polynomials identically zero
I'll propose two algorithms.
SAT
Every binary multivariate polynomial $f$ can be represented in the form
$$f(x) = \sum_{S \in \mathcal{F}} \prod_{i \in S} x_i,$$
where $\mathcal{F}$ has one set per monomial of $f$, representing the variables included in that monomial.  A multivariate function is identically zero iff all monomials have coefficient 0.  So, we can express your problem as follows.
Let us represent the partial assignment by sets $M$,  where $M$ indicates the set of variables that are are assigned and $v$ their values, and $\overline{M} = \{1,2,\dots,100\}\setminus M$ the complement of $M$.  Then the condition is that after we make the partial assignment and simplify by collecting monomials, all monomials must have coefficient zero.  This can be represented as
$$\Phi(M,v) = \bigwedge_{R \subseteq \overline{M}} \bigg[ \sum_{\substack{S \in F,\\S\cap \overline{M}=R}} \prod_{i \in S \cap M} v_i = 0 \bigg].$$
That's a huge logical formula, but it can be equivalently expressed with fewer clauses as
$$\Phi(M,v) = \bigwedge_{S \in \mathcal{F}} \bigg[ \sum_{\substack{T \in \mathcal{F},\\S\cap \overline{M}=T \cap \overline{M}}} \prod_{i \in S \cap M} v_i = 0 \bigg].$$
If $f$ has $n$ monomials, $\Psi(M,v)$ can thus be expressed as a sum of $n^2$ products of subsets of $v$.  This can be expressed as a SAT formula of size $O(n^2)$, by introducing boolean variables to represent $M,v$ and applying the Tseitin transform.
Now your problem can be expressed as checking whether there exist $M,v$ such that
$$\Phi_1(M,v) \land \dots \land \Phi_{100}(M,v) \land |M| \le k,$$
where $\Phi_i(M,v)$ is the formula corresponding to requiring $f_i$ be identically zero.  This can be checked with a SAT solver, and you can use binary search over $k$ to find the smallest $k$ for which this is satisfiable.
Unfortunately, if each $f_i$ has $n$ monomials, I suspect this will lead to a SAT instance with a small constant times $5000n^2$ CNF clauses, which sounds awfully large for existing SAT solvers to handle.
There are some optimizations that might make this more practical:

If we know that we are looking for an assignment to $k$ variables, the size of the formula for $\Phi(M,v)$ can be further reduced by additionally adding the constraint $|S \triangle T|\le k$ to the $T$'s we sum over, where $\triangle$ denotes the symmetric difference of sets.)

It might be helpful to use CryptoMiniSAT or some other SAT solver with built-in support for XOR (as each sum is a large XOR of boolean variables).


I'm not sure whether this will be enough to make your problem practically solvable in this way, though.
2QBF
We can formulate your problem as an instance of 2QBF as
$$\exists m, v . \text{wt}(m) \le k \land \forall w . f_1(m \;?\; v : w)=0 \land \dots \land f_{50}(m \;?\; v : w)=0,$$
where $m,v,w$ range over $\{0,1\}^{100}$, $k$ is a parameter, $\text{wt}(m)$ represents the Hamming weight of $m$ (the number of 1's in it), $m$ is a mask indicating which variables are assigned, $v$ indicates the values of the assigned variables, $w$ represents an assignment to the other variables, and
$$(m \;?\; v : w)_i = \begin{cases}v_i &amp;\text{if }m_i=1\\
w_i &amp;\text{if }m_i=0,\end{cases}$$
i.e., $m \;?\; v : w = (m \land v) \lor (\neg m \land w)$.  If you can test satisfiability of this kind of 2QBF formula, you can use binary search over $k$ to find the minimum number of variables to assign to.
So, you could try applying standard algorithms for 2QBF to your problem.  See https://cstheory.stackexchange.com/q/11022/5038.  I don't know whether to expect that to be successful -- the problem size might be a bit large for this to work.
2QBF is harder than SAT and unlikely to be solvable with multiple queries to a SAT solver (see Is 2QBF in P^NP?).