Yes, there is. The determinant of a matrix $A$ is the (signed) volume of the parallelopiped which is bounded the vectors whose columns are $A$.

Let's first consider the 2D case, and then extend to $n$-d.

$$
A = \begin{bmatrix} v_0 & w_0 \\ v_1 & w_1 \end{bmatrix} \\
|A| = v_0 w_1 - v_1w_0
$$
Think of $\hat v = (v_0, v_1)$ and $\hat w = (w_0, w_1)$. Now, the cross product of $v$ and $w$, which is

$$
|v \times w| = |v| \cdot |w| \sin(\theta)
$$

which is the area of the parallelogram with adjacent sides of length $|v|$ and $|w|$, with the sides having an angle of $\theta$ between them.

This immediately gives us some nice geometric understanding of determinant properties:

a determinant is 0 if the volume generated by the vectors is 0. One way for this to happen is to pick two of the same vectors. Hence, there is $0$ volume "between" the two vectors

The Jacobian (if you've studied multivariable calculus) gives you the "local change of volume" which we need to scale the integral correctly. Hence, the Jacobian pops up as an integration factor

A determinant is maximum if all your column vectors are orthogonal to each other - that is, you have an eigenbasis (which diagonalized your matrix)