Let $n$ be a positive integer and $\epsilon>0$. Write down $n$ linear equations on the variables $a_1,\dots,a_n$ in the following way. First, pick $n$ independent linear combinations of $a_1,\dots,a_n$, where the coefficients are $\pm 1$ or $0$. Second, for each linear combination, set a required value.
The system of equations must have a unique solution. Does there exist $\delta>0$ (in terms of $\epsilon,n$) such that if all required values are in $[-\delta,\delta]$, then the unique solution has $|a_i|\leq\epsilon$ for all $i$?
Intuitively, perturbing the required values away from $0$ by not too much should not change the solution too much either. But how can we formally show it?
Unless I'm misunderstanding your question, what we really have here is a full-rank matrix $M$, and a fixed $\epsilon > 0$. You are asking if there exists $\delta > 0$ such that for any vector $b$ with $|b_j| < \delta$ for all $j$, there exists vector $a$ such that $Ma = b$ and $|a_j| < \epsilon$ for all $j$.
First, notice that the map $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ defined by $f(x) = Mx$ is a continuous linear bijection, since $M$ is a full-rank matrix. Now we make the following observation: let $B_\epsilon$ be the open cube $(-\epsilon,\epsilon)^n$. Consider the image of $B_\epsilon$ under $f$. Since $f$ is a linear surjection, it's an open map, making $f(B_\epsilon)$ open in $\mathbb{R}^n$. And since $f(0) = 0$, there is some $\delta >0$ such that $B_\delta \subset f(B_\epsilon)$.
Taking the $\delta$ above, we're done. For any $b \in B_\delta$, there exists a unique $a \in B_\epsilon$ such that $Ma = b$. Notice that your additional assumption that $M$ is a matrix of only $0$s and $1$s was unnecessary.