Is there a deeper meaning behind the "determinant" formula for the cross product? We all know that for all vectors $\mathbf{a}, \mathbf{b} \in \mathbb{R^3}$, if $(a_x,a_y,a_z)^\top$ is the component form of $\mathbf{a}$ and similarly $(b_x, b_y, b_z)^\top$ is the component form of $\mathbf{b}$ then, the cross product can be evaluated by the formal determinant
\begin{equation}
\mathbf{a} \times \mathbf{b} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 
 a_x & a_y & a_z \\ 
b_x & b_y & b_z 
\end{vmatrix}
\end{equation}
However, I'm not satisfied with such an understanding. I think the formula is too elegant to be a mere formal, memorisation aid. So is there a generalisation of the determinant, which has an obvious geometric meaning, the same way the determinant can be interpreted the area multiplication factor of a linear transformation?
In geometric algebra, one can generalise the determinant of the matrix using the wedge product. If $\mathbf{a,b,c}$ are vectors in $\mathbf{R}^3$ then the determinant of the matrix, which has the vectors $\mathbf{a,b,c}$ as columns is $|\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}|.$ However, the wedge product is defined for general multivectors, so I tried evaluating 
$(\mathbf{ii} + a_x\mathbf{j} + b_x\mathbf{k}) \wedge (\mathbf{ji} + a_y\mathbf{j} + b_y\mathbf{k}) \wedge (\mathbf{ki} + a_z\mathbf{j} + b_z\mathbf{k})$ to obtain,
$\left(a_y b_z-a_z b_y\right) \mathbf{jk} +  \left(a_y-b_z\right) \mathbf{ijk}.$ However, the result doesn't have any obvious connection to the cross product.

The coordinate expansion of a wedge product is
$$\begin{aligned}\mathbf{a} \wedge \mathbf{b}
&= (\mathbf{e}_i a_i) \wedge (\mathbf{e}_j b_j) \\
&= \sum_{i, j} a_i b_j \mathbf{e}_i \wedge \mathbf{e}_j \\
&= \sum_{i < j} \begin{vmatrix}a_i & a_j \\ b_i & b_j\end{vmatrix}\mathbf{e}_i \wedge \mathbf{e}_j.\end{aligned}
$$
In $\mathbb{R}^3$, this can be related to the wedge product by a duality transformation (i.e. factoring out of a pseudoscalar for the space which such as $ \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 $), as in
$$\begin{aligned}\mathbf{a} \wedge \mathbf{b}
&=\begin{vmatrix}a_1 & a_2 \\ b_1 & b_2\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2+\begin{vmatrix}a_2 & a_3 \\ b_2 & b_3\end{vmatrix}\mathbf{e}_2 \mathbf{e}_3+\begin{vmatrix}a_3 & a_1 \\ b_3 & b_1\end{vmatrix}\mathbf{e}_3 \mathbf{e}_1 \\
&=\mathbf{e}_3\begin{vmatrix}a_1 & a_2 \\ b_1 & b_2\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3+\mathbf{e}_1\begin{vmatrix}a_2 & a_3 \\ b_2 & b_3\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3+\mathbf{e}_2\begin{vmatrix}a_3 & a_1 \\ b_3 & b_1\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \\
&=(\mathbf{a} \times \mathbf{b}) \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3.\end{aligned}
$$
The above factorization has made use of $ \mathbf{e}_i \mathbf{e}_i = 1 $, and $\mathbf{e}_i \mathbf{e}_j = -\mathbf{e}_j \mathbf{e}_i, i \ne j $.
Note that the wedge product has the geometric meaning of the area of the parallelopiped that are spanned by the two vectors, except that it is an oriented area, with a bivector orientation that equals the geometric product (or wedge product) of two perpendicular vectors that lie in the span of the plane.  
In 3D you can have a mapping between the normal to that oriented area (i.e. the cross product).  Which normal you get depends on the choice of the pseudoscalar used in the duality transformation.
The rationale for your choice of that multivector wedge product is not clear, so it is not suprising that it does not look anything like the cross product.
