This is a sequel to this question, though none of the background there should be necessary. Yly's excellent answer to that question gives a few different ways of stating the below question for $n = 2$, which may be helpful.
Short version: Consider a set of positive integers $A =\{m_i\}_{i=1}^n$ with $m_i < m_{i+1}$ and $\mathrm{gcd}(A) = 1$. Form an infinite binary sequence $s$ with $x$ $1$s and arbitrarily many $0$s subject to the following constraint: If there is a $0$ in position $j$, i.e., $s_j = 0$, then $s_{j+m_i} = 0$ for all $i$. Let the first entry of $s$ be $s_0$, and set $s_0 = 0$. How many such sequences are there for fixed $x$ and $\{m_i\}$?
More details: It quickly follows that the positions of the entries in $s$ which must be $0$s are the elements of a numerical semigroup, i.e., an infinite set of nonnegative integers containing $0$ that is closed under addition. (Wikipedia link here.) If $S$ is a semigroup, it has a minimal set of generators, i.e., the smallest set of relatively prime non-negative integers $\{m_i\}$ such that every element in $S$ can be written in the form $\sum_i a_i m_i$ for non-negative $a_i$. By removing superfluous $m_i$ if necessary, we can take $A$ from above to be the minimal set of generators of the semigroup $S$ that gives the positions of the required $0$s in the sequence. The positive integers not contained in $S$ are called the gaps of $S$, denoted $G(S)$. The number of gaps, called the genus of $S$, is denoted $g(S)$. For our sequence, the gaps are the possible positions of the $1$s.
A few results quickly follow from this picture. If for fixed $\{m_i\}$, we let $N_0(x)$ be the number of sequences beginning with a $0$ and with $x$ $1$s, then $N_0(x = g(S)) = 1$, when all the gaps are filled, and $N_0(x) = 0$ for $x > g(S)$. It can also be shown that $N_0(1) = m_1 - 1$, as only the gaps before the first generator can be filled by a solitary 1, since if $s_i = 0$ for $i = 0, 1, \ldots, m_1$, then $s_i = 0$ for all $i$.
However, finding $N_0(x)$ for $1 < x < g(S)$ seems much more difficult. For example, Let $A = \{3,7\}$. We see that $G(S) = \{1,2,4,5,8,11\}$ and $g(S) = 6$. Let $x = 2$. Clearly we can take $s_1 = s_2 = 1$. We see that if we fill the gap at position 1 with a $0$, i.e. set $s_1 = 0$, then $s_4 = s_8 = s_{11} = 0$ as well. It follows that if $s_1 = 0$, we must have $s_2 = s_5 = 1$. If we instead take $s_1 = 1$ and $s_2 = 0$, then $s_5 = s_8 = s_{11} = 0$, so we can either take $s_2 = 1$ or $s_4 = 1$. Note that setting $s_2 = 0$ in the latter case does not force $s_4 = 0$. Thus we find $N_0(2) = 3$, with the gaps filled with $1$s and the actual sequences being $$(1,2) \text{ and } (1,4) \text{ and } (2,5), \qquad 011 \text{ and } 01001 \text{ and } 001001,$$ with the infinite trailing $0$s suppressed. We can do similar computations to find $N_0(x) = 2, 2, 1$ for $x = 3, 4,$ and $5$, respectively. The gaps filled with $1$s and the sequences are $$\begin{matrix} x = 3: & (1,2,4) \text{ and } (1,2,5) & 01101 \text{ and } 011001\\ x = 4: & (1,2,4,5) \text{ and } (1,2,5,8) & 011011 \text{ and } 011001001\\ x = 5: & (1,2,4,5,8) & 011011001\\ \end{matrix}.$$
It is unclear how to systematize this method. Clearly some understanding of how the gaps of $S$ "knock out" some of the subsequent gaps when the entries of $s$ corresponding to the gaps are set to $0$ is necessary. It seems like thinking about the residues of the gaps modulo the $m_i$ might be useful, but I've been unable to make much progress for general $\{m_i\}$.
There is reason to believe (see Yly's answer to the question linked at the top) that the solution involves counting integer partitions subject to various restrictions, so a closed form for $N_0(x)$ is likely impossible, but a generating function or an expression in terms of objects derived from $S$ would still be useful.