How generalize of inner product for three vectors? For two vectors $A,B\in R^n$, the inner product of them is
$$
A\cdot B = |A||B|\cos\theta
$$
where $\theta$ is the angle of $A,B$.
Its geometric meaning is the projection of $A$ multiply by $B$. In my view, it can be treated as a invariant of parallelogram generated by $A,B$, since the project of $A$ multiply by $B$   is equal to   the project of $B$ multiply by $A$. Just like the area is the invariant of  parallelogram, since it is base times height no matter which side is choosed as base.
As we know , there is n-dimensional volume, when $n=2$, it is area. The n-dimensional volume also can be treated as invariant of parallel polyhedra. But seemly, there is not "n-dimensional" inner product, which also be invariant of parallel polyhedra from its geometric meaning, and be inner product when $n=2$. I want such an inner product. From the geometry, I try some, liking
$$
I(A,B,C) = |A||B||C|\cos AB\cos BC
$$
where $AB$ is the angle of $A,B$. But obviously,
$$
I(A,B,C) \ne I(B,A,C).
$$
So, I ask here, whether there is  generalization of inner product which can act on three vectors? (I don't care about the inner of function.)

In fact, I feel  the inner product is very strange. Although I know its practical meaning,  I feel I never understand it. I feel it can be treated as   dual of volume in some sense (just feel).
Maybe, this is a stupid problem. Just like asking how divide three numbers to make it liking multiply three numbers. But I am not sure, so ask here. Thanks for any help.

(2021/10/22)  For making  the problem more precise, I add some my think. In fact, it should be a multilinear map
$$
I: R^n\times R^n\times R^n \rightarrow R
$$
satisfy :  $\forall X,Y,Z\in R^n$
(1) $ I(X,Y,Z) = I(X,Z,Y)= I(Y,X,Z)= I(Y, Z,X) = I(Z,X,Y)= I(Z,Y,X) $
(2) $I(X,X,X)=0 \iff X=\theta$.
(3) $I$ is multilinear.
(4) $X\bot Y, Y\bot Z, Z\bot X$, $I(X,Y,Z)=0$.
(5) It doesn't depend on the choice of basis.
Obviously, the  first three is  the generalization of  inner product. The (5) is to avoid some  misunderstand. About (4), I am not sure it should be instead by $X\bot Y \Rightarrow I(X,Y,Z)=0$. I feel this is too strong. So I use (4).
I try to deal it from the tensor. Assuming
$e_i$ is basis of $R^n$ and $\omega^i$ is the dual basis of $e_i$, then,
$$
I=I_{ijk}\omega^i\otimes \omega^j\otimes \omega^k
$$
Therefore, the problem is transfered to the existence of $I_{ijk}$ satisfy (1) (2) (4). But I fail to know whether it is existence. In fact, uniqueness is also unknown.

(2021/10/23) According to   Jyrki's hint, I delete $I(X,X,X)\ge 0$.
 A: There is a symmetric multilinear form that takes in $n$ vectors and spits out a number:
\begin{align*}
\mathbb{R}^n \times \mathbb{R}^n \times \ldots \times \mathbb{R}^n & \to \mathbb{R} \\
\left(\begin{pmatrix}v_1^1 \\ v_2^1 \\ \vdots \\ v_n^1\end{pmatrix} , \begin{pmatrix}v_1^2 \\ v_2^2 \\ \vdots \\ v_n^2\end{pmatrix} , \ldots  , \begin{pmatrix}v_1^n \\ v_2^n \\ \vdots \\ v_n^n \end{pmatrix}  \right) &\mapsto \sum_{i=1}^n v_i^1 v_i^2 \cdots v_i^n 
\end{align*}
Symmetric means that if you switch the place of two of the input vectors, the result does not change. Multilinear means that it is linear in every entry (you can take out addition and scalar multiplication).
This is a generalization of the standard bilinear form $\left\langle \begin{pmatrix} v_1 \\ v_2\end{pmatrix} , \begin{pmatrix} w_1 \\ w_2\end{pmatrix} \right\rangle = v_1w_1 + v_2w_2$. I am not sure about it's geometric interpretation though.
A: The requirement that $I(X,X,X)$ never vanishes non-trivially is impossible to achieve if we assume multilinearity and full symmetry $(1)$. Assuming $n\ge2$.
Let $X$ and $Y$ be two linearly independent vectors. Fix them for now. Let $u$ and $v$ range over the reals. Then
$$
f(u,v):=I(uX+vY,uX+vY,uX+vY)= u^3I(X,X,X)+3u^2vI(X,X,Y)+3uv^2 I(X,Y,Y)+v^3I(Y,Y,Y).
$$
If $(2)$ holds, the coefficients of $u^3$ and $v^3$  above are non-zero. It follows that $f(u,v)$ takes both positive and negative values at some points $P=(u_+,v_+)$ and $N=(u_-,v_-)$. But $f(u,v)$ is a polynomial, so it is continuous. Therefore it takes the value zero along any path (on the $uv$-plane) connecting the points $P$ and $N$. Therefore $f(u,v)$ vanishes also at points other than $u=v=0$.

Come to think of it, the above argument does not need symmetry (item $(1)$) at all. Multilinearity and $(2)$ are incompatible requirements.
