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 Cayley-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.

Remark 2. As I understand Liviu Nicolaescu's suggestion in comments, let $\varphi:\mathbb{R}^n\to E$ be the linear map that sends every standard base element $e_k$ into $v_k$. Let $\hat{\mu}_n$ be the pull-back of $\mu_n$ with respect to $\varphi$. Let $B=\varphi^{-1}(B_{E})$, and let $\Omega_n$ be the volume of the standard Euclidean ball in $\mathbb{R}^n$. Then, $\hat{\mu}_n(B)=\Omega_n$ (as was proved by Busemann), and so the constant factor from the first remark can be obtained by comparing $\Omega_n$ with the actual volume of $B$. Moreover, since the pull-back of the parallelepiped is the unit cube, we get that $$\mu_n(P)=\hat{\mu}_n([0,1]^n)=\lambda_n([0,1]^n)\frac{\hat{\mu}_n(B)}{\lambda_n(B)}=1\cdot\frac{\hat{\mu}_n(B)}{\Omega_n}=\frac{\hat{\mu}_n(B)}{\Omega_n},$$ where $\lambda_n$ is the Lebesgue measure on $\mathbb{R}^n$. Thus, the question is reduced to the following:

Let $E$ be a real normed space, and let $v_1,...,v_n\in E$ be linearly independent. Let $\lambda_n$ be the Lebesgue measure on $\mathbb{R}^n$, and let $\varphi:\mathbb{R}^n\to E$ be the linear map with $\varphi(e_k)=v_k$, $k\in\overline{1,n}$. How to calculate $\lambda_n[\varphi^{-1}( B_{E})]$?

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 Cayley-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.

Remark 2. As I understand Liviu Nicolaescu's suggestion in comments, let $\varphi:\mathbb{R}^n\to E$ be the linear map that sends every standard base element $e_k$ into $v_k$. Let $\hat{\mu}_n$ be the pull-back of $\mu_n$ with respect to $\varphi$. Let $B=\varphi^{-1}(B_{E})$, and let $\Omega_n$ be the volume of the standard Euclidean ball in $\mathbb{R}^n$. Then, $\hat{\mu}_n(B)=\Omega_n$ (as was proved by Busemann$^1$), and so the constant factor from the first remark can be obtained by comparing $\Omega_n$ with the actual volume of $B$. Moreover, since the pull-back of the parallelepiped is the unit cube, we get that $$\mu_n(P)=\hat{\mu}_n([0,1]^n)=\lambda_n([0,1]^n)\frac{\hat{\mu}_n(B)}{\lambda_n(B)}=1\cdot\frac{\hat{\mu}_n(B)}{\Omega_n}=\frac{\hat{\mu}_n(B)}{\Omega_n},$$ where $\lambda_n$ is the Lebesgue measure on $\mathbb{R}^n$. Thus, the question is reduced to the following:

Let $E$ be a real normed space, and let $v_1,...,v_n\in E$ be linearly independent. Let $\lambda_n$ be the Lebesgue measure on $\mathbb{R}^n$, and let $\varphi:\mathbb{R}^n\to E$ be the linear map with $\varphi(e_k)=v_k$, $k\in\overline{1,n}$. How to calculate $\lambda_n[\varphi^{-1}( B_{E})]$?

1 - see Theorem 3.7.5 in Minkowski Geometry by A.C.Thompson.

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.

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 Cayley-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.

Remark 2. As I understand Liviu Nicolaescu's suggestion in comments, let $\varphi:\mathbb{R}^n\to E$ be the linear map that sends every standard base element $e_k$ into $v_k$. Let $\hat{\mu}_n$ be the pull-back of $\mu_n$ with respect to $\varphi$. Let $B=\varphi^{-1}(B_{E})$, and let $\Omega_n$ be the volume of the standard Euclidean ball in $\mathbb{R}^n$. Then, $\hat{\mu}_n(B)=\Omega_n$ (as was proved by Busemann), and so the constant factor from the first remark can be obtained by comparing $\Omega_n$ with the actual volume of $B$. Moreover, since the pull-back of the parallelepiped is the unit cube, we get that $$\mu_n(P)=\hat{\mu}_n([0,1]^n)=\lambda_n([0,1]^n)\frac{\hat{\mu}_n(B)}{\lambda_n(B)}=1\cdot\frac{\hat{\mu}_n(B)}{\Omega_n}=\frac{\hat{\mu}_n(B)}{\Omega_n},$$ where $\lambda_n$ is the Lebesgue measure on $\mathbb{R}^n$. Thus, the question is reduced to the following:

Let $E$ be a real normed space, and let $v_1,...,v_n\in E$ be linearly independent. Let $\lambda_n$ be the Lebesgue measure on $\mathbb{R}^n$, and let $\varphi:\mathbb{R}^n\to E$ be the linear map with $\varphi(e_k)=v_k$, $k\in\overline{1,n}$. How to calculate $\lambda_n[\varphi^{-1}( B_{E})]$?