Is there a simple generalization of the OLS "weighted average of the slopes" interpretation of the slope coefficient to multivariate predictors? The slope coefficient  of univariate (single predictor, single response) OLS can be written as a weighted average of the pairwise slopes between $(x,y)$ pairs, specifically:
$$ \hat{\beta} = \frac{\sum_{i\neq j}(y_{i}-y_{j})(x_{i}-x_{j})}{\sum_{i\neq j}(x_{i}-x_{j})^2}\\
= \sum_{i\neq j}W_{ij}\frac{(x_{i}-x_{j})}{(y_{i}-y_{j})} \text{ where } W_{ij} := \frac{(y_{i} - y_{j})^2}{\sum_{i\neq j}(x_{i}-x_{j})^2}$$
Let $\|X\|$ denote the Euclidean norm of $X$. Is the generalization to multivariate OLS obtained by replacing $(x_{i}-x_{j})^2$ everywhere in the above with $\|x_{i}-x_{j}\|^2$ true?
I can obtain something that looks like it could at least potentially be put into such a form (with some extra work) In the following way: Plug $\hat{\beta} = (X^{T}X)^{-1}X^{T}Y$ back into the OLS objective function, diagonalize the positive definite matrix $X^{T}X = UDU^{T}$ and make the change of variables $Z = XU^{T}$. Then expanding the resulting equivalent problem back into iterated summations looks suspiciously like it could work, but I don't see quite how to finish up.
I strongly suspect that I'm missing some easy trick that would let me make the last steps. Or$\ldots$ is it just not true?

No, but something similar works.
In the one-slope case you need two points to define a line.  With $p$ parameters to estimate you need $p+1$ points.
So, for any subset of $p+1$ points you get a unique $p$-plane, which defines a set of coefficient estimates, and you take a weighted average of these to get the OLS estimate. The weight needs to be (proportional to) the squared volume of the simplex (hypertriangle) formed by the points (or, equivalently, proportional to the squared determinant of the design matrix for the subset of points)
Lots of people have independently come up with this idea (including me), but a published source is https://www.jstor.org/stable/25050821
