No, there is no result in this form because in dimension 3 or higher it is allowed to have some non-first minima relatively small even when the first minimum is very small.

For example, for any $N>0 $ consider the lattice $$ \Lambda_N = \frac 1 N \mathbb Z \times \mathbb Z \times \mathbb Z^{n-3}\times 2N\mathbb Z$$ with unit determinant. Then the set $$ S_N = \left\{\left(\frac k N,1 , 0,\ldots, 0\right):\ -N\leq k\leq N\right\}$$ is a set of $2N+1$ primitive vectors of $\Lambda_N$ contained in the unit hypercube.

Now, to answer your question, take $B=N$ prime and the lattices $N\Lambda_N\subset \mathbb Z^n$ to see that there is no such constant $C_n$.

No, there is no result in this form because in dimension 3 or higher it is allowed to have some non-first minima relatively small even when the first minimum is very small.

For example, for any $N>0 $ consider the lattice $$ \Lambda_N = \frac 1 N \mathbb Z \times \mathbb Z \times \mathbb Z^{n-3}\times N\mathbb Z$$ with unit determinant. Then the set $$ S_N = \left\{\left(\frac k N,1 , 0,\ldots, 0\right):\ -N\leq k\leq N\right\}$$ is a set of $2N+1$ primitive vectors of $\Lambda_N$ contained in the unit hypercube.

Now, to answer your question, take $B=N$ prime and the lattices $N\Lambda_N\subset \mathbb Z^n$ to see that there is no such constant $C_n$.

No, the determinant datum is not enough to give such result, because in dimension 3 or higher it is allowed to have some non-first minima relatively small even when the first minimum is very small.

For example, for any $N>0 $ consider the lattice $$ \Lambda_N = \frac 1 N \mathbb Z \times \mathbb Z \times \mathbb Z^{n-3}\times 2N\mathbb Z$$ with unit determinant. Then the set $$ S_N = \left\{\left(\frac k N,1 , 0,\ldots, 0\right):\ -N\leq k\leq N\right\}$$ is a set of $2N+1$ primitive vectors of $\Lambda_N$ contained in the unit hypercube.

Now, to answer your question, take $B=N$ prime and the lattices $N\Lambda_N\subset \mathbb Z^n$ to see that there is no such constant $C_n$.

No, there is no result in this form because in dimension 3 or higher it is allowed to have some non-first minima relatively small even when the first minimum is very small.

For example, for any $N>0 $ consider the lattice $$ \Lambda_N = \frac 1 N \mathbb Z \times \mathbb Z \times \mathbb Z^{n-3}\times 2N\mathbb Z$$ with unit determinant. Then the set $$ S_N = \left\{\left(\frac k N,1 , 0,\ldots, 0\right):\ -N\leq k\leq N\right\}$$ is a set of $2N+1$ primitive vectors of $\Lambda_N$ contained in the unit hypercube.

Now, to answer your question, take $B=N$ prime and the lattices $N\Lambda_N\subset \mathbb Z^n$ to see that there is no such constant $C_n$.