Let ##\vec{b}_i## (##i \in \{1,2,3 \}##) denote any basis of ##\mathbb{R}^3## (it can also be a basis defined locally by generalized coordinates) and ##\vec{e}_i## arbitrary Cartesian coordinates. Further define the coordinates of the position vector as
$$\vec{x}=\tilde{x}^i \vec{b}_i=x^j \vec{e}_j.$$
Now we need a manifestly covariant description of the volume element. In Cartesian coordinates we simply have
$$\mathrm{d}V=\mathrm{d}^3 \vec{x}=\mathrm{d} x^1 \mathrm{d} x^2 \mathrm{d} x^3.$$
The trick is to rewrite this as
$$\mathrm{d}^3 \vec{x}=\Delta_{jkl} \mathrm{d} x^j \mathrm{d} x^k \mathrm{d} x^l,$$
where ##\Delta_{jkl}## is the Levi-Civita symbol, which is totally antisymmetric under interchange of its indices and ##\Delta_{123}=1##.
Unfortunately the Levi-Civita symbol is not a general tensor, because defining the transformation matrix between the general and the Cartesial coordinates as
$${T^j}_k=\frac{\partial x^j}{\partial \tilde{x}_k}$$
we have
$$\mathrm{d}^3 \vec{x}=\Delta_{jkl} {T^{j}}_{j'} {T^{k}}_{k'} {T^{l}}_{l'} \mathrm{d} \tilde{x}^{j'} \mathrm{d} \tilde{x}^{k'} \mathrm{d} \tilde{x}^{l'}=\det \hat{T} \mathrm{d} \Delta_{j'k'l'} \tilde{x}^{j'} \mathrm{d} \tilde{x}^{k'} \mathrm{d} \tilde{x}^{l'},$$
i.e., you get the Jacobian of the transformation in addition, as it must be for the proper transformation of the volume element. I assume that this determinant is positive (i.e., that the orientation of the basis ##\vec{b}_j## is the same as that of the Cartesian basis ##\vec{e}_j##, which you can always get by choosing the right order of the basis vectors).
Now we have a Euclidean space, i.e., there is the scalar product. In Cartesian components it's simply represented by ##\delta_{jk}##, i.e., you have for any two vectors ##\vec{x}## and ##\vec{y}##
$$\vec{x} \cdot \vec{y} = \delta_{jk} x^j y^k=\delta_{jk} {T^{j}}_{j'} {T^{k}}_{k'} \tilde{x}^j \tilde{y}^k := g_{j'k'} \tilde{x}^j \tilde{y}^k.$$
In matrix-vector notation you thus have
$$\hat{g}=\hat{T}^t \hat{T} \; \Rightarrow g=\det{\hat{g}}=(\det \hat{T})^2.$$
So you can write the volume element as
$$\mathrm{d}^3 \vec{x}=\det T \Delta_{j'k'l'} \mathrm{d} \tilde{x}^{j'} \mathrm{d} \tilde{x}^{k'} \mathrm{d} \tilde{x}^{l'} = \sqrt{g} \Delta_{j'k'l'} \mathrm{d} \tilde{x}^{j'} \mathrm{d} \tilde{x}^{k'} \mathrm{d} \tilde{x}^{l'}.$$
Since for Cartesian coordinates ##g_{jk}=\delta_{jk}## we can define the Levi-Civita tensor components as an invariant tensor via
$$\epsilon_{jkl}=\sqrt{g} \Delta_{jkl}.$$
Then the general covariant definition of the volume element in terms of the position-vector components reads
$$\mathrm{d}^3 \vec{x} =\epsilon_{jkl} \mathrm{d} \tilde{x}^j \mathrm{d} \tilde{x}^k \mathrm{d} \tilde{x}^l.$$
Now if you have a scalar density (like mass density ##\rho(\vec{x})## in non-relativistic fluid dynamics), a mass element in a small volume around ##\vec{x}## is given by [corrected according to #3]
$$\mathrm{d} m=\mathrm{d}^3 \vec{x} \rho(\vec{x})=\sqrt{g} \rho \mathrm{d} \tilde{x}^1 \mathrm{d} \tilde{x}^2 \mathrm{d} \tilde{x}^3.$$
For the mass-current-density then you have [corrected according to #3]
$$\mathrm{d} m v^i= \sqrt{g} \rho v^i \mathrm{d} \tilde{x}^1 \mathrm{d} \tilde{x}^2 \mathrm{d} \tilde{x}^3.$$