My answer is pretty much taken from Winitzki's Linear Algebra via Exterior Products, a very good book available legitimately for free online.
Here's the idea. Let's say we have two vectors, $\mathbf{a}, \mathbf{b} \in \mathbb{R}^2$. It's not hard to show that the area of the parallelogram with vertices $\mathbf{0}, \mathbf{a}, \mathbf{b}$ and $\mathbf{a} + \mathbf{b}$ is $|\mathbf{a}| \cdot |\mathbf{b}| \sin\theta$, where the angle between our two vectors is $\theta$.
Let's give this function a name; $Ar(\mathbf{a}, \mathbf{b})$. As we demand linearity, then it must be the case that
\begin{align*}Ar(\mathbf{a + b},\mathbf{a + b}) &= Ar(\mathbf{a} , \mathbf{a}) +Ar(\mathbf{a} , \mathbf{b}) +Ar(\mathbf{b} , \mathbf{a}) + Ar(\mathbf{b} , \mathbf{b})\\
&= 0 + Ar(\mathbf{a} , \mathbf{b}) + Ar(\mathbf{b} , \mathbf{a}) + 0\\
&= 0,\end{align*}
where all the zeros come from the fact that the area $Ar(\mathbf{x} , \mathbf{x})$ can only sensibly be $0$ for all vectors $\mathbf{x}$.
Thus, we're forced to set $Ar(\mathbf{b} , \mathbf{a}) = -Ar(\mathbf{a} , \mathbf{b})$, in order to get the sane result that $Ar(\mathbf{a+b} , \mathbf{a+b}) = 0$.
This isn't the whole story, of course, but that's the idea: vanishing on a linearly dependent input list and (multi)linearity force us to use antisymmetry. In order to get sensible results in terms of volumes of parallelopipeds, we need oriented volume.
That, in my opinion, is the best motivator for the determinant: thinking in terms of signed volumes. It has a geometric intuition and even 'detects' linear dependence, like we saw above. It provides a great way to see why antisymmetry is essentially required.