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The conjecture seems to fail for $n=3$ and $(q_1,q_2,q_3) = (9,10,10)$, when ${\rm vol}(E) = (2^3/3!) 9 \cdot 10 \cdot 10 = 1200$ but $ \#(E \cap {\bf Z}^3) = 1199$. In general it seems (and is probably not hard to prove) that if $(q_1,q_2,q_3) = (k,k+1,k+1)$ then $$ \#(E \cap {\bf Z}^3) - {\rm vol}(E) = -\frac23 k^2 + \frac{16}{3} k + 5 $$ which is negative for $k \geq 9$, and increasingly so as $k$ grows. Other examples with $n=3$ and $q_1 \leq q_2 \leq q_3 \leq 20$ are $(6,17,18)$, $(7,13,14)$, $(8,15,16)$, $(9,17,18)$, $(9,19,19)$, $(10,19,20)$ with a shortfall of $-1$, $-17/3$, $-17$, $-31$, $-1$, $-143/3$ respectively, and suggesting similar generalizations.