(If you're not familiar with differential forms, this is a good opportunity to learn about them, since they are exactly the appropriate objects to use in the answer to your question. This may seem a bit technical, but once you are familiar with differential forms, it should make perfect sense.)
Let's say we're talking about a volume-preserving flow, generated by a vector field V on Euclidean space
ℝn. Let ω denote the volume form.
That V preserves ω means that the Lie derivative L
Vω = 0.
In other words,
d(ιVω) + ιV(dω) = 0,
where ι
V denotes the contraction of a differential form by V.
Since ω is already an n-dimensional form on
ℝn, we have
dω = 0
automatically, which means that
d(ιVω) = 0
Now the flux form τ, i.e., the (n-1)-form that must be integrated over an (n-1)-dimensional manifold M to get the flux of V through M, is given by just contracting ω by V:
τ = ιVω
and so from the previous equation, the exterior derivative of the flux form vanishes everywhere:
dτ = 0.
In Euclidean space (and other spaces with sufficiently simple topology), when the exterior derivative d of any differential form σ vanishes everywhere, this implies by the de Rham theorem that σ itself is the exterior derivative of some form of one lower dimension. Hence we have that there exists an (n-2)-dimensional form μ such that
τ = dμ.
Now, the general version of Stokes's theorem that works in any dimension states that for a compact manifold M and a differential form ρ of dimension = dim(M) - 1, we have
∫M dρ = ∫∂M ρ.
(In fact, all the "usual vector calculus integration theorems" you refer to can be shown to be special cases of this general Stokes's theorem for differential forms.)
Applying this Stokes's theorem to the flux form τ with
τ = dμ,
we have
∫M τ = ∫M dμ
and so we get
flux through M = ∫M τ = ∫∂M μ.
This is exactly what you asked about: How to calculate the flux through an (n-1)-dimensional manifold by integrating something else over its boundary.
