symplectic, I think you missed my point. The convention does not matter. The RHS has 1-tensor coefficients, and the LHS has a 1-form coefficient. The only reason that I mentioned the convention was to clarify that the position of the index tells you that it is a 1-form.
Here's a different (more explicit) way to look at it. A 3-surface is basically a volume. A differential "area" of the 3-surface would have the form dxdydz, right. So, generalize these differentials to 1-tensors: dxαdyβdzγ, where these differentials are some specific directions in space-time. This product has units of 3-surface area (volume), but it is a 3-tensor, not a 1-form. You convert it to a 1-form by contracting with the Levi-Civita symbol. This is similar to converting a 1-tensor to a 1-form by contracting with the metric.
I don't know if this is standard notation, but, instead of using indices, imagine that the Levi-Civita symbol is a function that "eats" n-tensors and spits out (4-n)-forms.
ε(⋅,⋅,⋅,⋅)
So, in this case, the Levi-Civita symbol "eats" three 1-tensors, dx, dy, and dz.
ε(⋅,dx,dy,dz)=~dV
The tilde in front is my way to identify the object as a n-form. 1 argument of the Levi-Civita symbol is left open, so the result is a 1-form.
This is similar to the idea of the metric "eating" an n-tensor, resulting in a (2-n)-form.
η(⋅,⋅)
So, if the metric "eats" a vector (1-tensor), the result is a 1-form.
η(⋅,dx)=~dx
In the index approach, this is called "lowering" the index.
One more point to make: ε(dt,dx,dy,dz) is Lorentz invariant. This is true regardless of what dt, dx, dy, and dz represent, as long as they are 1-tensors.