I hope the computations don't obscure the intuition here. $C(n,d)/M(n,d)$ is a conditional probability. Imagine rolling $6$ dice and getting the most common total, $21$. The conditional probability that each number came up exactly once is small, but it wouldn't be terribly shocking for this to be the case. Suppose you roll $6000$ dice and get a total of $21000$. Now, what's the conditional probability that exactly $1000$ of the dice showed a $1$, ... and exactly $1000$ showed $6$? It would be much more of a surprise, and the conditional probability isn't bounded away from $0$ as the number of dice increases. That said, here are some computations:

For each fixed $n$, you can use the Central Limit Theorem and log-convexity to obtain the asymptotics of $M(n,d)$ (Brendan McKay pointed out that unimodality is not enough to get an upper bound on $M(n,d)$). The coefficients of $(\frac{1+x+...+x^n}{n+1})^d$ are approximately normal. The standard deviation is $\sigma = \sqrt{d (n^2+2n)/12}$, and $\frac{M(n,d)}{(n+1)^d} \approx \frac {1}{\sqrt{2 \pi}}\sigma^{-1}$ since the peak of the density of the standard normal distribution is $\frac{1}{\sqrt{2\pi}}$. So, for fixed $n$,

$$M(n,d) \approx (n+1)^d \sqrt{\frac{6}{\pi d(n^2+2n)}}.$$

Another application of the Central Limit Theorem would also approximate $C(n,d)$ but Stirling's formula is simpler. All powers of $e$ and most powers of $d$ cancel.

$${d \choose \frac {d}{n+1},\frac{d}{n+1},...\frac{d}{n+1}} = \frac{d!}{{\frac {d}{n+1}}!^{n+1}}$$

$$\approx \sqrt{2\pi d} \bigg/ \bigg[\bigg(\frac{1}{n+1}\bigg)^d \sqrt{2 \pi \frac{d}{n+1}}^{n+1}\bigg]$$

$$\approx (n+1)^{d+(n+1)/2} (2\pi d)^{-n/2} $$.

When $d$ is large but not divisible by $n+1$ the most even integral division is not much different.

The quotient $M(n,d)/C(n,d)$ is proportional to some function of $n$ times $d^{(n-1)/2},$ so there is no upper bound for the quotient only depending on $n$, and you can't bound the conditional probability $C(n,d)/M(n,d)$ away from $0$ regardless of $d$.