Work modulo $3$ with $3 \equiv 0$ and $2 \equiv -1.$ The seven index triples $\{i, i+1, i+3\}$ form a Fano plane: every pair of positions lies in exactly one triple.

Classify by the set $Z$ of positions holding a $3:$ a product survives only if its triple misses $Z,$ and each surviving product is $\pm 1$ according to how many $2$’s its triple holds.

Work modulo $3:$ entries $3$ are $0$ and entries $1, 2$ are $\pm 1.$ Because the differences of $\{0, 1, 3\}$ hit every nonzero residue mod $7$ exactly once, the seven triples $\{i, i+1, i+3\}$ are the lines of a Fano plane on the positions: every pair of positions lies on exactly one line, and any two lines meet in exactly one point. Let $Z$ be the set of positions holding a $3$ and $k = |Z|.$ A product term survives exactly when its line avoids $Z.$ If $t_L$ is the number of entries equal to $2$ on that line, its contribution is $(-1)^{t_L},$ and the linear condition constrains the $7 - k$ values $\pm 1$ to sum to $0$ mod $3.$

Casework on $k.$ $k = 7:$ the all-$3$s tuple works: $1.$ $k = 6:$ a single $\pm 1$ can’t sum to $0:$ none. $k = 5:$ no line survives; the two nonzero entries must be a $1$ and a $2:$ $\binom{7}{2} \cdot 2 = 42.$ $k = 4:$ three $\pm 1$s sum to $0$ only if all equal, and the three nonzero positions must not form a line, else its product is $\pm 1:$ $(35 - 7) \cdot 2 = 56.$ $k = 3:$ four $\pm 1$s must split two and two; exactly one line avoids a non-line $Z$ (spoiling the sum), while a line $Z$ is avoided by no line: $7 \cdot \binom{4}{2} = 42.$ $k = 2:$ five $\pm 1$s must go four and one; exactly two lines avoid $Z,$ meeting at a point $p$ and covering the five positions, and their products cancel exactly when the lone minority value avoids $p:$ $\binom{7}{2} \cdot 2 \cdot 4 = 168.$ $k = 1:$ six $\pm 1$s sum to $0$ if all equal or three of each; the four lines avoiding $Z$ pairwise meet in the six nonzero positions, and since the product of all four line-products is $+1,$ we need exactly two negative lines. All-equal gives $0$ or $4$ negative lines; for three $2$’s, viewing positions as edges of $K_4$ on the four lines, a line is negative exactly when it has odd degree in the chosen $3$-edge set, and exactly the $12$ three-edge paths (of the $\binom{6}{3} = 20$ subsets) give two odd degrees: $7 \cdot 12 = 84.$ $k = 0:$ seven $\pm 1$s need two or five $2$’s, which make $4$ or $3$ lines negative respectively, but $7 - 2t \equiv 0 \pmod 3$ needs $t \equiv 2 \pmod 3:$ none.

The total is $1 + 42 + 56$ $+ 42 + 168 + 84 = 393.$