Let's say we have a smooth, orientable, Riemannian manifold $M$ with metric $g$, where $dim(M)=n$. Let $U$ be a submanifold contained in M such that $dim(U)=a$, where $1\leq a\leq n$
Let $x^1,\dots,x^n$ be a set of local coordinate functions of $M$, and let $u^1,\dots,u^n$ be another.
Then the $a$-dimensional volume of $U$ in the $x^n$ coordinate system is: $$\int_U \omega_a$$
Where, in local coordinates: $$\omega_a=\sqrt{|det(g)|}dx^1\wedge\dots\wedge dx^a$$
Now, let $\mu_a$ be $\omega_a$ in the $u^n$ coordinate system. Then:
$$\mu_a = det\left(\frac{\partial (x^1,\dots,x^a)}{\partial (u^1,\dots,u^a)} \right) \sqrt{|det(g)|} du^1\wedge\dots\wedge du^a$$
Thus, the $a$-dimensional volume of $U$ in the $u^n$ coordinate system is:$$\int_U \mu_a$$
I've heard that the volume of $U$ is dependent on our choice of coordinates, to the extent that it can be non-zero in one coordinate frame and zero in another. However, this doesn't make sense to me, since the only way this could happen is if the Jacobian of the coordinate change is equal to zero (afaik), which wouldn't make much sense.
So, I guess my question is this: is the volume of a submanifold dependent on our choice of local coordinates and, if so, is there a coordinate free formulation of submanifold volume?
Any Riemannian manifold has a volume computed in terms of the metric, which is coordinate invariant. A submanifold acquires a Riemannian metric by restriction/pullback, so in this way it has a coordinate-invariant notion of volume.
However, this definition of the volume for $U$ is not equivalent to the one you have mentioned. This is because it defines volume as the integral of a certain multiple of $dx^1\wedge \cdots \wedge dx^a$, but the $x^i$ are not necessarily coordinates for $U$ - they could even be orthogonal to $U$ (in which case this "volume" will be zero).
In other words, if one of the 1-forms $dx^i$ is orthogonal to the subspace $T^*U \subset T^*M$ everywhere on $U$ then the differential form $dx^1\wedge\cdots\wedge dx^a$ will vanish along $U$, and the integral $\int_U \omega_a$ will vanish as well. If on the other hand, all of the $dx^i$ are parallel with $T^*U$ then you will compute the correct version of volume. If the $dx^i$ are neither parallel nor orthogonal with $U$, generically you will get an incorrect but non-zero answer.