Let $E$ be a real normed space, and let $v_1,...,v_n\in E$ be linearly independent. The parallelepiped defined by these vectors is $P=\{\sum_{i=1}^{n}\alpha_i v_i|~0\le\alpha_i\le 1\}$. Since $E$ is a metric space, we can define the $n$-dimensional Hausdorff measure $\mu_n$ on it.

Is there a method to calculate $\mu_n(P)$ if we know "everything" about $v_1,...,v_n$ and $E$?

If the norm of $E$ is induced by an inner product, then the volume can be calculated by taking the square root of the Gram matrix of $v_1,...,v_n$. I am hoping there is something of the kind if $E$ is a concrete space of functions or sequences. Since the $(i,j)$ entry of the Gram matrix is $<v_i,v_j>=\frac{1}{2}\left(\|v_i+v_j\|^2-\|v_i\|^2-\|v_j\|^2\right)$, maybe one can get $\mu_n(P)$ from the distances inside the parallelepiped somehow (also Kelley-Menger determinant comes to mind).

Remark. If we identify $\mathbb{R}^n$ with $span (v_1,...,v_n)$ via a linear isomorphism, the pull-back of $\mu_{n}$ will be an invariant measure, and so a multiple of the usual Lebesgue measure. The only unknown thing is the constant factor.