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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 $\{ \bar{x} \in V : \| \bar{x} \| < (j+1)U \}$ 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.

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.

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 $\Lambda_j=\mathbb{Z}v_1+\dots+\mathbb{Z}v_j$, which is a lattice in $V_j=\mathbb{R}v_1+\dots+\mathbb{R}v_j$. By Minkowski's second theorem, the fundamental parallelepiped $P_j$ of $\Lambda_j$ in $V_j$ has $j$-volume $\gg_j\lambda_1\dots\lambda_j$, and its translates by the elements of $\{ \bar{x} \in \Lambda : \| \bar{x} \| < U \}$ are pairwise disjoint and they all lie in $\{ \bar{x} \in V : \| \bar{x} \| < (j+1)U \}$ whose $j$-volume is $\ll_j U^j$. This proves the upper bound.

Added. Per request of the OP, I give more detail regarding the last paragraph. The $j$-volume $\text{vol}_j$ in $V_j$ is the volume determined by the euclidean norm in $V_j$ inherited from $\mathbb{R}^n$, i.e. an orthonormal basis in $V_j$ spans a $j$-cube of $j$-volume $1$. By Minkowski's second theorem applied to $\Lambda_j$ and $V_j$, $$\text{vol}_j(P_j)\geq 2^{-j}\cdot\lambda_1\dots\lambda_j\cdot\text{vol}_j(B_j),$$ 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 $1$, and so $\text{vol}_j(B_j)\gg_j 1$. Combining with the previous inequality, we get $\text{vol}_j(P_j)\gg_j\lambda_1\dots\lambda_j$ as claimed originally.

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 $\{ \bar{x} \in V : \| \bar{x} \| < (j+1)U \}$ 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.

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 $\mathbb{Z}v_1+\dots+\mathbb{Z}v_j$, which is a lattice in $V=\mathbb{R}v_1+\dots+\mathbb{R}v_j$. By Minkowski's second theorem, the fundamental parallelepiped of this lattice in $V$ has $j$-volume $\gg_j\lambda_1\dots\lambda_j$, and its translates by the elements of $\{ \bar{x} \in \Lambda : \| \bar{x} \| < U \}$ are pairwise disjoint and they all lie in $\{ \bar{x} \in V : \| \bar{x} \| < (j+1)U \}$ whose $j$-volume is $\ll_j U^j$. This proves the upper bound.

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 $\Lambda_j=\mathbb{Z}v_1+\dots+\mathbb{Z}v_j$, which is a lattice in $V_j=\mathbb{R}v_1+\dots+\mathbb{R}v_j$. By Minkowski's second theorem, the fundamental parallelepiped $P_j$ of $\Lambda_j$ in $V_j$ has $j$-volume $\gg_j\lambda_1\dots\lambda_j$, and its translates by the elements of $\{ \bar{x} \in \Lambda : \| \bar{x} \| < U \}$ are pairwise disjoint and they all lie in $\{ \bar{x} \in V : \| \bar{x} \| < (j+1)U \}$ whose $j$-volume is $\ll_j U^j$. This proves the upper bound.

Added. Per request of the OP, I give more detail regarding the last paragraph. The $j$-volume $\text{vol}_j$ in $V_j$ is the volume determined by the euclidean norm in $V_j$ inherited from $\mathbb{R}^n$, i.e. an orthonormal basis in $V_j$ spans a $j$-cube of $j$-volume $1$. By Minkowski's second theorem applied to $\Lambda_j$ and $V_j$, $$\text{vol}_j(P_j)\geq 2^{-j}\cdot\lambda_1\dots\lambda_j\cdot\text{vol}_j(B_j),$$ 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 $1$, and so $\text{vol}_j(B_j)\gg_j 1$. Combining with the previous inequality, we get $\text{vol}_j(P_j)\gg_j\lambda_1\dots\lambda_j$ as claimed originally.