The conjecture fails 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, 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. This was found experimentally, but is not hard to prove: the integer points are those for which $\left| x_1 \right| + \left| x_2 \right| + \left| x_3 \right| \leq k$ (i.e. the lattice points inside the $(k,k,k)$ octahedron), together with the $4(k+1)$ points with $x_1=0$ and $\left| x_2 \right| + \left| x_3 \right| = k+1$. By a standard induction there are $\frac43 k^3 + 2k^2 + \frac83 k + 1$ integer solutions of $\left| x_1 \right| + \left| x_2 \right| + \left| x_3 \right| \leq k$, etc.
The 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. Here's the gp code, which enumerates $E \cap {\bf Z}^3$ by setting up a generating function for the number of $(x_1,x_2,x_3)$ in each octahedral shell $\sum_{i=1}^3 \left| x_i \right| / q_i = n / (q_1 q_2 q_3).$
P(q,H) = sum(n=1,H\q,2*x^(q*n),1+O(x^(H+1)))
{
D3(q1,q2,q3, T) =
T = P(q1*q2, q1*q2*q3) * P(q1*q3, q1*q2*q3) * P(q2*q3, q1*q2*q3);
sum(n=0,q1*q2*q3,polcoeff(T,n)) - (4/3)*q1*q2*q3
}
B = 20
{
for(q1=1,B,for(q2=q1,B,for(q3=q2,B,
d = D3(q1,q2,q3);
if(d<=0, print([q1,q2,q3,d]))
)))
}