Let $s(m,r)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct residue classes modulo $m$ such that $a_1 + a_2 + a_3 + a_4 \equiv r \pmod{m}$. Because the total number of quadruples of distinct residue classes is $m(m-1)(m-2)(m-3)$, it follows that
$$\sum_{r=1}^{m} s(m, r) = m(m-1)(m-2)(m-3).$$
Note that the pairing $(a_1, a_2, a_3, a_4) \longleftrightarrow (a_1+1, a_2+1, a_3+1, a_4+1)$ provides a one-to-one correspondence between quadruples of distinct residue classes that sum to $r$ and quadruples of distinct residue classes that sum to $r+4$. Hence $s(m, r) = s(m, r+4)$, so
$$s(m, r) = s(m, r+4) = s(m, r+8) = \dots$$
Now $\gcd(4, m) = 1$ because $m$ is odd, so the numbers $r, r+4, r+8, \dots, r+4(m-1)$ form a complete residue system modulo $m$. Thus the values $s(m, r)$ are all equal. From the displayed summation above it follows that
$$s(m, r) = (m-1)(m-2)(m-3) = m^3 - 6m^2 + 11m - 6$$
for all $r$. Because $D(m) = s(m, 0)$, it follows that the required polynomial is $q(x) = x^3 - 6x^2 + 11x - 6$, and $c_1 = 11$.