## Upper bounds

This is not a full answer. But provides some ideas.

### Duality

First, I'd like to spell out the duality argument and show that if $\frac1p + \frac1q = 1$, then $C_p = C_q$.

For this we will need the well-known dual characterization of $L^p$ norms:
$$ \|f \|_p = \sup_{g: \|g\|_q = 1} \int fg $$
Now, on the left we want to estimate $\| f- \int f\|_p$, so we compute, for an arbitrary $g\in L^q$ with unit norm the quantity $\int (f - \int f)g$. Observe that since $f-\int f$ has zero mean, its integrals against constants vanish. So we have
$$ \int (f-\int f)g = \int (f-\int f)(g - \int g) $$
Now modifying $f$ using the same reason, we find
$$ = \int (f-c)(g-\int g) \leq \|f-c\|_p \|g - \int g\|_q $$
where we applied, for the inequality, Holder's inequality.

To conclude, we have that for any $f\in L^p$ and $g\in L^q$ that
$$ \int (f-\int f)g \leq \|f - c\|_p \|g - \int g\|_q \leq \|f - c\|_p \cdot C_q \|g - c'\|_q $$
for any $c'\in \mathbb{R}$.
This in particular applies to $c' = 0$.

Now optimizing over all $g$ with $\|g\|_q = 1$ you find
$$ \|f - \int f\|_p \leq C_q \|f - c\|_p $$
for every $f\in L^p$ and every $c\in \mathbb{R}$. This shows that $C_p \leq C_q$.

But the argument is symmetric in $p$ and $q$, thus we conclude that $C_p = C_q$.

### Interpolation bounds

First note that the question can be rephrased as looking for the operator norm of the linear operator $T: f \mapsto f - \int f$, since given $g = T(f)$, then $T^{-1}(g) = \{ f + c: c\in \mathbb{R}\}$. So the problem of finding the operator norm of $T$ can be written as finding

$$ \sup_{f\in L^p\setminus \{0\}} \frac{\|Tf\|_p}{\|f\|_p} =
\sup_{g\in L^p\setminus \{0\}, \int g = 0} \sup_{f \in T^{-1}(g)} \frac{\|g\|_p}{\|f\|_p} = \sup_{g\in L^p\setminus \{0\}, \int g = 0} \frac{\|g\|_p}{\inf_{c} \|g+c\|_p}$$

The fact that $C_p$ can be found explicitly for $p = 1, 2, \infty$ means that you can apply Riesz-Thorin-Stein interpolation.

For $p > 2$, we can write it as
$$ \frac1p = \frac{\theta}{2} + \frac{1-\theta}{\infty}$$
(or more simply, $\theta = 2/p$). Complex interpolation tells us that our operator having operator norm $C_2 = 1$ and $C_\infty = 2$ must have operator norm
$$ C_p \leq C_2^\theta C_\infty^{1-\theta} = 2^{1-\theta} = 2^{1-2/p}$$
In particular, $C_p < 2$ for all $p \neq 1,\infty$.

For comparison, here are the numerical values of the function above:

$p$ |
$2^{1-2/p}$ |

3 |
1.26 |

4 |
1.414 |

5 |
1.516 |

6 |
1.587 |

8 |
1.682 |

10 |
1.741 |

15 |
1.823 |

20 |
1.866 |

25 |
1.892 |

50 |
1.945 |

So you see that the numerical lower bounds are compatible with the theoretical upper bounds, but there is a gap.