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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) $2g-2 = 16-12 = 2$, so $g=3$, and since the matrix you provided is non-singular, $b_1(\Sigma(4,4,4)) = b_1(B) = 6$.