Your expectation is correct. Let $\lambda_{j} \leq U < \lambda_{j+1}$. There are independent lattice vectors $v_1,\dots,v_j\in\Lambda$ such that $\|v_i\|=\lambda_i$ for $1\leq i\leq j$.

Consider the linear combinations
$ c_1v_1+\dots +c_j v_j\in\Lambda$ with integral coefficients $c_i\in\mathbb{Z}$ satisfying $|c_i|\leq U/(2n\lambda_i)$. Their sup-norms are $\leq U/2<U$, and their number is $\gg_n U^j/(\lambda_1\dots\lambda_j)$. This proves the lower bound.

Now consider the subspace $V_j=\mathbb{R}v_1+\dots+\mathbb{R}v_j$ equipped with the $j$-volume coming from the euclidean norm in $\mathbb{R}^n$, i.e. an orthonormal basis in $V_j$ spans a $j$-cube of $j$-volume $1$. Then, $\Lambda_j=\Lambda\cap V_j$ is a lattice in $V_j$ with successive minima $\lambda_1,\dots,\lambda_j$, hence by Minkowski's second theorem its fundamental parallelepiped $P_j$ has $j$-volume
$$\text{vol}_j(P_j)\geq 2^{-j}\cdot\lambda_1\dots\lambda_j\cdot\text{vol}_j(B_j),$$
where $B_j=\{ \bar{x} \in V_j : \| \bar{x} \| \leq 1 \}.$
The sup-norm is majorized by the euclidean norm, hence $B_j$ contains a euclidean $j$-ball of radius $1$, which shows that $\text{vol}_j(P_j)\gg_j \lambda_1\dots\lambda_j$.
On the other hand,
$$\{ \bar{x} \in \Lambda : \| \bar{x} \| < U \}=\{ \bar{x} \in \Lambda_j : \| \bar{x} \| < U \},$$
whence the translates of $P_j$ by the elements of this set are pairwise disjoint and they all lie in $\{ \bar{x} \in V_j : \| \bar{x} \| < (j+1)U \}$ whose $j$-volume is $\ll_j U^j$. This proves the upper bound.

**Remark.** The above argument was enhanced and corrected by the OP's comments, for which I am grateful.