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]))
)))   
}

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

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.

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]))
)))   
}