Without saying "cross product" explain why there is a skew-symmetric angular momentum tensor In Space-Time-Matter Hermann Weyl claims that had we lived in a world of more than three spacial dimensions, we would have known all along that quantities such as angular momentum are skew-symmetric tensors, and the "error" of using the three-component "vector product" would never have occurred. Pages 66-67.
His first example of a skew-symmetrical(sic) tensor is angular momentum:

In mechanics the skew-symmetrical tensor product of two vectors occurs— 1. As moment of momentum (angular momentum) about a point $\mathcal{O}.$ If there is a point-mass at $\mathcal{P}$ and if $\xi^1,\xi^2,\xi^3$ are the components of $\vec{\mathcal{OP}}$ and $u^{i}$ are the (contra-variant) components of the velocity of the points at the moment under consideration, and m its mass, the momentum of momentum is defined by
$$L^{ik}=m\left(u^{i}\xi^{k}-u^{k}\xi^{i}\right).$$

Why/how should we define something called angular momentum as a skew-symmetric tensor without first learning about it as a cross-product?
I'm no expert in the application of her theorem, but it may be worth pointing out that (IIRC) Emmy Noether was Weyl's graduate student.
 A: TL;DR: OP's reference to Noether's theorem is exactly right. Angular momentum $x^{[i}p^{j]}$ is the Noether charge for rotational symmetry.

*

*The Noether charge in point mechanics reads
$$ Q~=~ p_i X^i - h T, \tag{1}$$
where $X^i$ and $T$ are spatial and temporal generators, and where $p_i=\frac{\partial L}{\partial v^i}$ is the momentum and $h=p_iv^i-L$ is the energy.


*The $n$-dimensional rotation group $SO(n)$ has a Lie algebra
$$so(n)~=~\{\omega\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \omega^T=-\omega\} \tag{2}$$
consisting of antisymmetric matrices. It generates infinitesimal rotations.


*An infinitesimal spatial rotation
$$ \delta x^i ~=~\epsilon X^i, \qquad \delta t~=~\epsilon T, \tag{3}$$
(where $\epsilon$ is an infinitesimal parameter),
has generators$^1$
$$ X^i = \omega^i{}_jx^j, \qquad T~=~0.\tag{4}$$


*So the corresponding Noether charge is
$$ Q~\stackrel{(1)+(4)}{=}~p_i \omega^i{}_jx^j~=~\underbrace{x^{[j}p^{i]}}_{\text{ang. mom.}}\omega_{ij}.\tag{5} $$
See also e.g. this related Phys.SE post.
--
$^1$ We raise and lower spatial indices with the spatial metric $\delta_{ij}$.
A: Because the physical interpretation and expression are still the same, but if we are not in three dimensions it makes no sense to try to write it as a cross-product.
Thinking about higher dimensions is hard, so instead let us gain some intuition on two-dimensional mechanics. Following the same reasonings of usual 3D mechanics, but now restricted to a plane, we'll eventually find out that the quantity
$$L = m (xv_y - yv_x ) \tag{1}$$
is a conserved quantity in the absence of a torque. However, notice we can't write it as a cross-product: $L$ is a number, while $\mathbf{r} = (x,y)^{\intercal}$ and $\mathbf{v} = (v_x,v_y)^{\intercal}$ are vectors. $L$ is just a sort of weird combination that we could write as in Eq. (1).
Given that, suppose now that we lived in 4D. Eventually, we would pay attention to 2D problems (perhaps when answering a question in Phys.SE and afraid of tackling too many dimensions at once lol) and notice that for any pair of coordinates $r^i$ that we pick, there is a quantity $L^{ij}$ which is conserved when there is no torque "in that direction". Naturally, we eventually notice that all of these quantities are just the same notion in different directions and decide to simply write
$$L^{ij} = m (r^i v^j - r^j v^i).$$
We'd notice that we are repeating ourselves and defining $4^2 = 16$ objects instead of the $6$ we needed, but that is okay: it is easier to just arrange them in that way.
In 3D, we do exactly the same thing, the only difference being that we notice there is a simpler way of writing these objects. Instead of writing $L^{ij}$ all the time and having to deal with all those redundant entries, we can use the Levi-Civita symbol to define a different object $L^i$ (with only one index) such that
$$L^i = \frac{1}{2} \epsilon^{ijk} L^{jk}.$$
This definition yields $L^1 = L^{23}$, $L^{2} = L^{31}$, and $L^{3} = L^{12}$, and one can show that $L^i$ transforms as a vector under rotations. Hence, in many applications it might be easier to simply define a new operation between vectors that take $\mathbf{r}$ and $m\mathbf{v}$ straight to the new vector we've just build, $\mathbf{L}$. We could perhaps denote it by a cross and write $\mathbf{L} = m \ \mathbf{x \times v}$. We'll notice that this quantity doesn't always behave as a "true" vector, specially when considering its behavior when taking its mirror image, so sometimes we might want to remember this and call it a pseudovector. Nevertheless, in most cases, we'll get away by treating it as a regular vector, as long as we stick to three dimensions.
A: Rotation is more intimately related to notions of area and planes than it is related to length or lines. Consider Kepler's second law, which says that the line between a planet in orbit and the focus of the orbit sweeps out area at a constant rate. If you calculate the rate of sweeping out of area, you find that it is proportional to the magnitude of the angular momentum. So we should interpret the angular momentum as a rate of area-sweeping (in general, it needs to be weighted by mass). As this is a geometric area we should be biased against using a vector for it, as vectors are geometrically lines.
Further, note that rotational motion naturally takes place in a plane, specifically the plane spanned by the velocity and the position. In the planet example, this plane, in which the planet and star lie, is also conserved. The orientation of that plane in space is in fact the other part of the conserved quantity of angular momentum, in addition to the magnitude from above. Again, a vector is a line-like thing, not a plane-like thing, so it's a bad choice for representing the orientation of angular momentum. (In 3D, note that the vector representation of angular momentum chooses a vector perpendicular to the relevant plane; i.e. the vector is exactly in the direction the rotation isn't.)
There's also a problem with choosing the direction of a vector from a plane orientation. In 3D, you have to arbitrarily choose either the left-hand or right-hand rule. But intuitively physics (or at least classical physics) doesn't differentiate your hands, so why should our representation of physics in mathematics need this asymmetry? In 2D, there is no vector perpendicular to "the" plane. And in 4 and higher dimensions, there doesn't even exist a viable left-/right-hand rule. But in all of these settings, the idea of rotation still makes sense. If you just give up your insistence on treating angular momentum as a "line-like" vector, and treat it as a new "plane-like" object, perhaps all of these problems go away? (NB: as we live in a 3+1 dimensional universe, worrying about 4D rotations is a practical concern!)
So we come to the question of how to represent the combination of a plane orientation and a magnitude as an object we will call a bivector (in the same way a vector is the combination of a line orientation and magnitude). If two orthonormal vectors $\mathbf{\hat u},\mathbf{\hat v}$ span a plane, let's say $\mathbf{\hat u}\wedge\mathbf{\hat v}$ is the bivector representing that plane, where the wedge $\wedge$ is a new symbol we've just made up. We want the order to matter, as rotating from $\mathbf{\hat u}$ into $\mathbf{\hat v}$ is exactly the opposite of rotating from $\mathbf{\hat v}$ into $\mathbf{\hat u},$ so let's establish $\mathbf{\hat u}\wedge\mathbf{\hat v}=-\mathbf{\hat v}\wedge\mathbf{\hat u},$ and let's decide $\mathbf{\hat u}\wedge\mathbf{\hat v}$ is oriented in the same sense as the rotation from $\mathbf{\hat u}$ into $\mathbf{\hat v}$ (the short way). This also gives $\mathbf{\hat u}\wedge\mathbf{\hat u}=0,$ which reproduces the geometric statement that parallel vectors do not have any plane "between" them. Finally, let's make $\wedge$ bilinear. Given a basis for our ordinary vector space, this means the wedge of any two vectors can be decomposed into a sum of wedges between the basis vectors, with some coefficients. I.e. the wedges between the basis vectors create the basis bivectors. In 2D the basis vectors $\mathbf{\hat e}_1,\mathbf{\hat e}_2$ produce one basis bivector $\mathbf{\hat e}_1\wedge\mathbf{\hat e}_2.$ In 3D, the one new basis vector $\mathbf{\hat e}_3$ produces two new basis bivectors $\mathbf{\hat e}_1\wedge\mathbf{\hat e}_3,\mathbf{\hat e}_2\wedge\mathbf{\hat e}_3.$ In 4D we get 3 more bivectors, etc. Your cited formula for angular momentum corresponds to $\mathbf L=m\mathbf r\wedge\mathbf v:$ (twice) the rate of area ($\wedge$) swept ($\mathbf v$) by the line between an object and a point ($\mathbf r$), weighted by mass ($m$).
Finally, the conversion of $\mathbf L$ into an antisymmetric tensor comes from wanting to calculate its components (coefficients on the basis bivectors) in the same way we split vectors into components. Each basis bivector was made by wedging two of the basis vectors, so if we label each basis vector with one index, the basis bivectors should have two indices: $\mathbf{\hat e}_{ij}=\mathbf{\hat e}_i\wedge\mathbf{\hat e}_j.$ Similarly, each of the angular momentum bivector's components is identified by which basis bivector it is taken along, so the components are labelled with two indices as $L^{ij}$ and constitute $\mathbf L$ by $$\mathbf L=\sum_{\text{$\mathbf{\hat e}_{ij}$ in the basis}}L^{ij}\mathbf{\hat e}_{ij}=\frac12\sum_{i,j}L^{ij}\mathbf{\hat e}_{ij}.$$
There are multiple ways to take the sum on the left. I.e. if we choose to put $L^{ij}\mathbf{\hat e}_{ij}$ in the sum, we must not include $L^{ji}\mathbf{\hat e}_{ji}$ since $\mathbf{\hat e}_{ij},\mathbf{\hat e}_{ji}$ cannot be in a basis together, or we could choose the other way, but either choice should give the same result. But $\mathbf{\hat e}_{ij}=-\mathbf{\hat e}_{ji},$ so $L^{ij}=-L^{ji},$ meaning $L^{ij}$ comes out as an antisymmetric tensor.
So $L^{ij}$ is an antisymmetric tensor because it is the component representation of a bivector $\mathbf L,$ and the basis bivectors themselves have a certain antisymmetry. The existence and properties of bivectors and the identification of angular momentum as such an object are justified geometrically. In 3D, you can mistakenly identify bivectors with vectors because there happen to be the same number of basis bivectors as vectors. In other dimensionalities this will not happen, and you'll have to do it right.
A: Angular momentum is primarily an interesting concept because central forces conserve it. Since$$\frac{d}{dt}(u^i\xi^k)=\dot{u}^i\xi^k+u^iu^k$$becomes symmetric if the force $m\xi^k=\lambda u^k$ for some $\lambda$, in that case $L^{ik}:=m(u^i\xi^k-u^k\xi^i)$ is conserved.
A: 3D case
the angular momentum $~\mathbf L$ must be perpendicular to the vectors
$~\mathbf u~$ and  $~\mathbf \xi$
hence
$$\mathbf L\cdot \mathbf u=0\\
\mathbf L\cdot \mathbf \xi=0$$
those are two equations for the three unknows $~L_1~,L_2~,L_3$
you obtain the solution
$$\mathbf L= \left[ \begin {array}{c} {\frac {L_{{3}} \left( u_{{2}}\xi_{{3}}-u_{{
3}}\xi_{{2}} \right) }{u_{{1}}\xi_{{2}}-\xi_{{1}}u_{{2}}}}
\\  -{\frac {L_{{3}} \left( u_{{1}}\xi_{{3}}-\xi_{{1}
}u_{{3}} \right) }{u_{{1}}\xi_{{2}}-\xi_{{1}}u_{{2}}}}
\\  L_{{3}}\end {array} \right] 
$$
we can now choose arbitrary value for $~L_3$ but to make  the  angular momentum anti symmetric we choose
$$L_3=u_{{1}}\xi_{{2}}-\xi_{{1}}u_{{2}}$$
and obtain
$$\mathbf L=\left[ \begin {array}{c} u_{{2}}\xi_{{3}}-u_{{3}}\xi_{{2}}
\\  -u_{{1}}\xi_{{3}}+\xi_{{1}}u_{{3}}
\\  u_{{1}}\xi_{{2}}-\xi_{{1}}u_{{2}}\end {array}
 \right] 
$$
