The key is pointed out by HJRW in his comment: there's a missing piece in your explanation, which is the genus of the base $B$ of the Seifert fibration. Némethi writes: $$ 2g-2 = (n−2)A/a−\sum q_i, $$ which in your case gives (modulo mistakes in my computation) $2g-2 = 16 - 12 = 4$, so $g=6/2=3$, and since the matrix you provided is non-singular, $b_1(\Sigma(4,4,4)) = b_1(B) = 6$.
By the way, that the genus is 3 makes sense to me, since $\Sigma(4,4,4)$ is the 4-fold cyclic cover of $S^3$ branched over the link of the curve singularity of $C = \{x^4 + y^4 = 0\} \subset \mathbb{C}^2$ at the origin. To give an explicit description of its cover, it suffices to blow up the plane at the origin and observe that $\Sigma(4,4,4)$ is the boundary of cover of the disc bundle with Euler number -1 over the 2-sphere (morally, a compact version of the total space of the blow-up at the origin) branched over four fibres (the strict transform of $C$). This new 4-manifold is again a bundle, but now the base is the 4-fold cover of the 2-sphere branched over 4-points (which is indeed a genus-3 surface) and whose Euler number is $-4$.
