Distributions may be "integrated against" (i.e. evaluated on) test functions, the following notation (for $T$ a distribution) is fairly common: $$ T(f) = \int T(x) f(x)\, \mathrm{d}x $$ I'm wondering if anything is known about distributions which may be "integrated" in some sense over, say, smooth hypersurfaces (or other submanifolds) in the ambient space, so as to give a meaning to expressions of the form $$ \int_\Sigma T(x)\,\mathrm{d}x $$ for $\Sigma$ a smooth hypersurface. These objects would presumably exist "between" functions and distributions: not necessarily regular enough to have values at points but still regular enough that the smearing provided by test functions is not required in all directions.
I can only motivate this question by appealing to QFT: in the Wightman mindset, one would expect the energy-momentum tensor $T_{\mu\nu}(x)$ to be a distribution valued in some locally convex algebra of unbounded operators. Then there is a general principle that this should "generate" the momenta $P_\mu$ (the generators of the translations in the representation of the Poincaré group) in the sense that $$ P_\mu = \int_{x^0=0}T_{\mu 0}(0,{\bf x})\,\mathrm{d}^3{\bf x}. $$ Of course the formal expression on the right isn't immediately defined since the would-be integrand is a distribution. In fact this same problem threatens to arise whenever we have a conserved charge and try to write it as the integral of a current.
Moving to the simpler case where our distributions are scalar-valued (i.e. just standard distributions), we come back to the beginning of this question, wondering about integrals of distributions over hypersurfaces. Even though this seems like a fairly obvious issue, I have only seen it discussed in a paper by Verch, precisely in a QFT context.
Sounds like you want to work with what are sometimes called in mathematical circles differential forms that have distribution coefficients, also known more formally as currents. Steven Krantz and Harold Parks has an expository text: Geometric Integration Theory.