Formula for curved capacitance 'plates'
Oh, the formula is easy: it's
$$C = Q/V$$
and you find V by solving for the E field with a test charge +Q on one
plate, -Q on the other,
and integrating (line integral)
$$V = \int \overline E \cdot  d\overline L $$
from any convenient point on one electrode, to the other.

The charge isn't equally distributed over the area of the plates,
its location IS on the outer surface, with E inside the outer surface
set to zero.

It gets more complicated because of the dielectric and/or nearby metal
structures, and the geometry is not self-shielding, so technically
you need to know all the surroundings out to infinity to complete the
E-field calculation.

The electrode shapes are boundaries, with boundary conditions that
amount to the E-field component parallel to the surface vanishing at
zero distance from the surface.   As boundaries go, these are not
simple shapes.   The field solution, therefore, is not going to be
simple.

If the problem can be made two-dimensional (i.e. very long device,
with constant cross-section), there are conformal mapping solutions
that can create symmetry and make the on-paper problem tractable.
Otherwise, the use of finite-element modeling would give good results,
but not a general formula.   Practical capacitors have uniform field
in the dielectric, and negligible field outside the dielectric, so
approximate solutions are very simple.   This, however, is not a
practical capacitor in that sense.