In a Johnson counter the inverted output of the last flipflop is given as the input to the first one. So, when the output of the previous FF becomes $0,$ the current FF output becomes $0$ in the next clock cycle.
So, after initial value $1 = 0001,$ in the next clock cycle, all FF outputs will be $0 - 0000.$ Then the cycle goes like $1,0,8,12,14,15,7,3,1$
n-bit Johnson counter has 2n states. From initial state $7_{10}=0111_2$ it will go to $0011_2=3_{10}$ and then $0001_2=1_{10}$ followed by $0000_2=0_{10}$, $1000_2=8_{10}$, $1100_2=12_{10}$, $1110_2=14_{10}$ and $1111_2=15_{10}$ respectively.
