Solution:

Two overlapping triples of consecutive vertices share two labels, so their sums differ by labels three positions apart. Since all triple sums are multiples of $3,$ labels three apart are congruent mod $3.$ Thus the position classes $\{1, 4, 7\},$ $\{2, 5, 8\},$ $\{3, 6, 9\}$ each carry a single residue class of digits, and the digits $1$ through $9$ consist of exactly three digits from each residue class mod $3.$

Every triple of consecutive positions then contains one digit from each residue class, with sum $\equiv 0 + 1 + 2 \equiv 0 \pmod 3,$ so all $3!$ assignments of residue classes to position classes are acceptable, and within each position class the three digits can be arranged in $3!$ ways. That gives $3! \cdot (3!)^3 = 6^4 = 1296$ acceptable labelings.

Because the digits are distinct, no nontrivial rotation fixes a labeling, so the $1296$ labelings split into rotation classes of size $9,$ giving $\frac{1296}{9} = 144$ distinguishable arrangements.