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, all with two of the $q_i$
related by a simple ratio. The OP commented "That is quite surprising.
I assumed that is better to look for coprime $q_i$ for counterexamples",
and I confess that this surprised me too (though the proof for
$(k,k+1,k+1)$ explains the mechanism). Curiously the conjecture
seems to be true when the $q_i$ are pairwise coprime; in fact
an exhaustive search of the region $q_i \leq 256$ found no
counterexamples to the stronger inequality
$$
\#(E \cap {\bf Z}^3) - {\rm vol}(E) > q_1.
$$
It's not easy even to find examples where the two sides are nearly equal.
The smallest ratios that the search turned up all fit the pattern
$$
(q_1,q_2,q_3) = (2n^2-3n, 2n^2-2n-1, 2n^2-n-1)
$$
with
$$
\#(E \cap {\bf Z}^3) - {\rm vol}(E) = 2n^2-2n+5 = q_1 + O(q_1^{1/2})
$$
(any $n>1$). I have not tried to prove that this persists for all $n$,
though I guess it won't be too hard $-$ not as easy as the $(k,k+1,k+1)$
explanation but much easier than proving the inequality for all
pairwise coprime $q_1,q_2,q_3$.

Still, even in the pairwise coprime case one soon find
counterexamples with $n=3$ to the "Conjecture (strong version)"
that $E$ has at least as many lattice points than $E+x$ for any
$x \in {\bf R}^n$. We can even take $x = (\frac12, 0, 0)$,
and then test the conjecture by comparing the count of integer points
in the $(2q_1,q_2,q_3)$ octahedron with twice the $(q_1,q_2,q_3)$ count.
If the former is larger then we have a counterexample, and this happens
already for $(q_1,q_2,q_3) = (5,2,3)$, when $E$ has $49$ integral points
but $E+(\frac12,0,0)$ has $50$. There are $30$ more such examples
with $\max_i q_i \leq 16$. (As it happens $(2,3,5)$ is the $n=2$
case of $(2n^2-3n, 2n^2-2n-1, 2n^2-n-1)$, and larger $n$ account for
some of the other counterexamples.)

In the previous edit of this answer I gave some needlessly convoluted
**gp** code for computing the difference between the number $L$ of
integral points in $E$ and the volume $\frac43 q_1 q_2 q_3$ of $E$.
Here's a much simpler and faster program, which takes time about $q_1 q_2$
and negligible space, whereas the earlier code (using generating functions)
took time and space $q_1 q_2 q_3$.

```
{
L3(q1,q2,q3) =
sum(x1=0,q1,if(x1>0,2,1) *
sum(x2=0,q2+(-q2*x1)\q1,if(x2>0,2,1) *
(1 + 2*(q3+(-q3*(q2*x1+q1*x2))\(q1*q2)))
)
)
}
d3(q1,q2,q3) = L3(q1,q2,q3) - (4/3)*q1*q2*q3
```