My understanding is that you get a vector that points in the steepest ascending direction of a function $f(x_1,...x_n)$ at point $\hat{X}$ by computing the gradient $\nabla f(X_1, ... X_n)$.
Does this mean that the negative gradient $-\nabla f(X_1,...X_n)$ is the steepest descending direction of the function $f$ at that point? It is obvious to me that this is the case for the derivative of one variable, but I am not certain it is analogous to the gradient.
On
Yes it is, indeed recall that the directional derivative for every direction defined by a vector $\vec v$ by the doct product with the gradient, that is:
$$f_{\vec v}=\frac {\partial f}{\partial \vec v}=\nabla f \cdot \vec v=|\nabla f| \,|\vec v|\cos \theta$$
then the minimum is obtained when $\vec v$ is opposite to $\nabla f$ that is for $\theta=180°\implies \cos \theta=-1$.
Hint: the direction of steepest descent of $f$ at $(x_1, \ldots, x_n)$ is the direction of steepest ascent of $-f$ at the same point.