I think you may have forgotten that although $L$ and $C$ here are 'lumped' parameters that facilitate the abstract model of the transmission line (idealised by $LC$ elements per unit length), these modelling parameters must eventually be linked to physical reality, and therefore must necessarily include geometry.
Indeed, finding the relation between $L,C$ (and also $R,G$ in the lossy case) and the physical geometry of the line's (or waveguide's) cross section is a known classical problem and is the premise of microwave engineering and electromagnetic field theory for guided waves (see for example, Pozar, p.51 onwards, or Collin's books here and here), which are mainly applied electrodynamics.
Specifically, here are the general equations that link these parameters with geometry (arbitrary cross-section of a typical line that is longitudinally uniform) [see Pozar, p. 52 and figure 2.2]:
$$ L=\frac{\mu}{|I_{0}|^{2}}\int_{S} \bar{H}\cdot\bar{H}^{*} ds \ \ \ \ \ \ \ \text{H/m}$$
$$ C=\frac{\epsilon'}{|V_{0}|^{2}}\int_{S} \bar{E}\cdot\bar{E}^{*} ds \ \ \ \ \ \ \ \text{F/m}$$
$$ R=\frac{R_{s}}{|I_{0}|^{2}}\int_{C{1}+C_{2}} \bar{H}\cdot\bar{H}^{*} dl \ \ \ \ \ \ \ \Omega\text{/m}$$
$$ G=\frac{\omega\epsilon''}{|V_{0}|^{2}}\int_{S} \bar{E}\cdot\bar{E}^{*} ds \ \ \ \ \ \ \ \text{S/m}$$
where $S$ is the surface area of the cross section (arbitrary shape), $C_{1,2}$ are the inner and outer contours of that cross section, with $ds, dl$ being the respective surface and distance elements, $R_{s}$ being the surface resistance of any conductor (typically lining the outer/inner contours cylinders) with the dielectric matrial having imperfect (complex) permittivity $\epsilon=\epsilon'-i\epsilon''$, and $I_{0}, V_{0}$ are the (arbitrary) amplitudes of current or voltage of the travelling wave on the line.
As you can see from the integration over the contours and surfaces, and including the $E$ and $H$ fields within the structure, such calculations will inevitably depend on the geometry.
As a famous example, a coaxial line that has outer conductor of radius $b$, inner conductor radius $a$, with a dielectric sandwiched between these two conductive cylinders, with surface resistance $R_{s}$ for the conductor, will have (after performing the integrations above):
$$ L = \frac{\mu}{(2\pi)^{2}}\int^{2\pi}_{\phi=0}\int^{b}_{r=a}\frac{rdrd\phi}{r^{2}}=\frac{\mu}{2\pi}\ln(b/a)\ \ \ \ \text{H/m}$$
$$ C = \frac{\epsilon'}{(\ln b/a)^{2}}\int^{2\pi}_{\phi=0}\int^{b}_{r=a}\frac{rdrd\phi}{r^{2}}=\frac{2\pi\epsilon'}{\ln(b/a)}\ \ \ \ \text{F/m}$$
$$ G = \frac{\omega\epsilon''}{(\ln b/a)^{2}}\int^{2\pi}_{\phi=0}\int^{b}_{r=a}\frac{rdrd\phi}{r^{2}}=\frac{2\pi\omega\epsilon''}{\ln(b/a)}\ \ \ \ \text{S/m}$$
$$ R = \frac{R_{s}}{(2\pi)^{2}}\left[\int^{2\pi}_{\phi=0}\frac{ad\phi}{a^{2}}+\int^{2\pi}_{\phi=0}\frac{bd\phi}{b^{2}}\right]=\frac{R_{s}}{2\pi}\left(\frac{1}{a}+\frac{1}{b} \right)\ \ \ \ \Omega\text{/m}$$
As you can see, the result clearly depend on geomerty.
