The book Geometrical Vectors does exactly that. In fact, it considers the gradient vector to be a different kind of vector than a normal distance vector. If I have a vector between two points, and I compress space so that the points move closer together, then the distance vector gets smaller. However, the gradient vector gets bigger. The gradient is basically a density, it's units are units-of-whatever-your-taking-the-gradient-of / meters.
And, or course, the vector you get from a cross product only has 3 degrees of freedom because we live in a 3D world. The cross product of two 2D vectors is a scalar, the cross product of two 4D vectors needs 6 numbers to describe it. Even in 3D, if you reflect the original vectors in a mirror, the cross product now points the opposite direction, i.e. it depends on the handednes of the coordinate system.
That sounds like the beginnings of intuition for exterior algebra. While the gradient is a vector (i.e. it lives in a vector space), it's better thought of as being different to a standard 'displacement' vector.
Look at a topographical map of a landscape and note the contour lines. As you zoom into the contour lines (if they're detailed enough), they'll start to look more and more like parallel lines, densely spaced for a steep slope, or sparsely spaced for a mild slope. These parallel lines are the gradient.
A gradient and a vector 'fit together' to give a real number. The more parallel lines the vector pierces, the bigger the number [0]. So a gradient 'eats' a vector, spitting out a real number, and vice-versa. Just like a row vector 'eats' a column vector and spits out a real number.
I'm saying 'gradient' here, but really what I mean is 'one-form'. Language deliberately imprecise for all y'all mathematicians out there.
I'm not personally knowledgeable enough to suggest how this would behave in geometric algebra, I'm just smart enough to think that this would behave much cleaner in that framework.
Would really like to have that free time to delve and be able to suggest the alternative formulation myself.
And, or course, the vector you get from a cross product only has 3 degrees of freedom because we live in a 3D world. The cross product of two 2D vectors is a scalar, the cross product of two 4D vectors needs 6 numbers to describe it. Even in 3D, if you reflect the original vectors in a mirror, the cross product now points the opposite direction, i.e. it depends on the handednes of the coordinate system.