Lower Bound on the Volume of a Subset of a Convex Cone Generated by Algebraic Integers

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Let $K$ be a totally real number field that is Galois over $\mathbb{Q}$, with $[K:\mathbb{Q}]=n$. Let $\sigma_1,\dots,\sigma_n$ denote the embeddings from $K$ into $\mathbb{R}$, and denote by $\Sigma: K \to \mathbb{R}^n, \Sigma(x)=(\sigma_1(x),\dots,\sigma_n(x))$.

Consider a set of algebraic integers $x_1,\dots,x_t$ of $K$, and assume that $\Sigma(x_1^2),\dots,\Sigma(x_t^2)$ span $\mathbb{R}^n$. Now, consider the convex cone

$$ \mathcal{K}=\Big\{\sum_{i=1}^t\lambda_i\Sigma(x_i^2): \lambda_i \in \mathbb{R}^+\Big\}, $$

and the subset

$$ \mathcal{K}_1=\Big\{(v_1,\dots,v_n) \in \mathcal{K}: \prod_{i=1}^n v_i \leq 1\Big\}. $$

Question. Is it possible to determine a lower bound on the volume of $\mathcal{K}_1$ in terms of $x_1,\dots,x_t$ (or some properties of these generating integers) and some properties of $K$ (e.g. the discriminant, regulator etc)?

Edit: Perhaps I can add some context to this problem. A lattice is said to be perfect if it is uniquely determined (up to homothety) by its set of minimal vectors, and its shortest nonzero vector. It is known that the number of perfect lattices (up to homothety) in any given dimension is finite, and the convex cones generated by their minimal vectors are disjoint (see e.g. Perfect lattices in Euclidean Spaces, J. Martinet).

The notion of perfect lattices (or perfect quadratic forms) naturally translates to the algebraic setting. In my context, I am interested in the volume of the perfect cones for so-called "perfect unary forms" (see e.g. On additive generalisation of Voronoï’s theory of perfect forms over algebraic number fields, A. Leibak) intersected by the hyperplane $\big\{(x_1,\dots,x_n) \in \mathbb{R}^n: \prod_{i=1}^n x_i=1\big\}$, or in other words, the set of totally positive elements of $K_{\mathbb{R}}$ of algebraic norm no greater than $1$ that lie within a perfect cone.