Answer: $2n + 1$
Let $N = \overline{a_m \dots a_2 a_1}$ be a multiple of $d$ such that $a_i \in \{1, 2\}$ for every $1 \le i \le m$, where $d$ denotes $10^n - 7$. Since $N \ge 2d$ it is clear that $m \ge n + 1$. Suppose that $2n \ge m$. By setting $S = 7 \times \overline{a_m \dots a_{n+1}} + \overline{a_n \dots a_1}$ we get
$$S \le 7 \times \underbrace{22\dots2}_{n} + \underbrace{22\dots2}_{n} < 2d.$$
It follows from $d \mid N$ that $d \mid S$ and so we must have $S = d$ $\left(*)\right)$. But this equality is impossible. Indeed, by $\left(*)\right)$, we have $S \equiv 7a_{n+1} + a_1 \pmod{10}$ which implies $S \not\equiv 3 \pmod{10}$ since $a_1, a_{n+1} \in \{1, 2\}$. This contradicts to $\left(*)\right)$. Thus $m \ge 2n + 1$ and so, for our purpose, it suffices to show that $2\underbrace{1\dots11}_{n-1}\underbrace{2\dots22}_{n-2}111$ is divisible by $d$:

$$\begin{aligned} 2\underbrace{1\dots11}_{n-1}\underbrace{2\dots22}_{n-2}111 &= 2 \times 10^{2n} + \underbrace{1\dots11}_{n-1}2 \times 10^n + \underbrace{2\dots2}_{n-3}111 \\ &\equiv 2 \times 7^2 + 7 \times \underbrace{1\dots11}_{n-1}2 + \underbrace{2\dots2}_{n-3}111 \\ &\equiv \underbrace{7\dots7}_{n-2}84 + \underbrace{2\dots2}_{n-2}09 = d \equiv 0 \pmod{d}. \end{aligned}$$