Solution:

Work modulo $3.$ The value $3$ is the only multiple of $3,$ and each of the five products covers three cyclically consecutive positions, so if $x_i = 3$ exactly two products avoid position $i:$ those covering positions $i+1, i+2, i+3$ and $i+2, i+3, i+4$ (indices mod $5$). Their sum is $x_{i+2}x_{i+3}(x_{i+1} + x_{i+4}),$ and since $x_{i+2}x_{i+3}$ is not divisible by $3,$ the condition is $x_{i+1} + x_{i+4} \equiv 0 \pmod 3.$

Among the remaining values, $1$ and $4$ are $\equiv 1$ while $2$ and $5$ are $\equiv 2 \pmod 3,$ so positions $i+1$ and $i+4$ must take one value from each class: $2 \cdot 2 \cdot 2 = 8$ ordered choices. The other two values fill positions $i+2$ and $i+3$ in $2$ ways. With $5$ choices for the position of $3,$ the count is $5 \cdot 8 \cdot 2 = 80.$