A number $B$ having $n-1$ digits '1' and one digit '7' in decimal representation is of the form $B = A_n + 6 \cdot 10^k$ where $A_n$ is a number having $n$ digits '1', and $0 \le k < n$. Notice that if $3|n$ then the sum of the digits of $B$ is $3n+6$. Notice that
$$A_1 = 1,\ A_2 = 4,\ A_3 = 6,\ A_4 = 5,\ A_5 = 2,\ A_6 = 0 \pmod{7}$$
$$10^0 = 1;\ 10^1 = 3;\ 10^2 = 2;\ 10^3 = 6;\ 10^4 = 4;\ 10^5 = 5;\ 10^6 = 1 \pmod{7}$$
Let $n > 6$. Then $n = 6t + r$ where $t \ge 0$, $1 \le r \le 6$. Then
$$A_n = A_{6t+r} = A_{6t} + A_r \cdot 10^{6t} \equiv A_{6t} + A_r \equiv A_r \pmod{7}$$
If $6|n$ then $3|n$, so therefore we saw that $B$ is not prime in any case. Suppose $6$ does not divide $n$. Then we saw that $A_n \equiv A_r \pmod{7}$ and additionally $A_r$ is not congruent to $0$ modulo $7$. Put $A_r \equiv t \pmod{7}$. For this $t$, from the above it follows that we can choose $k$, $0 \le k \le 5$, such that $6 \cdot 10^k = -10^k \equiv -t \pmod{7}$. So this $B$ is divisible by $7$. For $n=6$ we saw that the numbers are composite. We need to check all cases $n \le 5$. For $n=5$
$$A_5 + 6 \cdot 10^2 = 2 - 2 = 0 \pmod{7}$$
For $n=4$, $1711 = 29 \cdot 59$. In the case $n=3$ we already saw that all numbers are composite. For $n=2$ and $n=1$ all numbers are prime.