Is this a known formula? $ \prod_{k=0}^n \left(1 - \frac{a_k}{N}\right)$

Let $f(x) = -\log(1-x)$ for $x \in [0,1)$. Since $f''(x) = \frac{1}{(1-x)^2} > 0$. $f(x)$ is convex over $[0,1)$.
By Jensen's inequality, for any $a_0,\ldots, a_n \ge 0$ such that $\sum_{i=0}^n a_i = N$, we have

$$-\sum_{i=0}^n\log\left(1 - \frac{a_i}{N}\right) \ge -(n+1)\log\left(1-\frac{1}{n+1}\right)$$ In the Taylor expansion of $f(x) = \sum\limits_{k=1}^\infty \frac{x^k}{k}$, all coefficients are positive. This leads to

$$-(n+1)\log\left(1-\frac{1}{n+1}\right) = (n+1)\sum_{k=1}^\infty \frac{(n+1)^{-k}}{k} > (n+1)\sum_{k=1}^1 \frac{(n+1)^{-k}}{k} = 1$$ As a result,

$$\prod_{i=0}^n\left(1 - \frac{a_i}{N}\right) \le \left(1-\frac{1}{n+1}\right)^{n+1} < e^{-1}$$ In short, $e^{-1}$ is indeed the upper limit for a finite $N$ for all partitions.

Since $$1-x\leq e^{-x},$$ if $x\geq 0,$ the upper bound you state: $$ \prod_{k=0}^n \left(1 - \frac{a_k}{N}\right)\leq e^{-a_1/N} \times \cdots\times e^{-a_N/N} =e^{-1}, $$ holds for any finite partition.

Your measure can be looked at as the $n^{th}$ power of the geometric mean of the relative sizes of the complements of partition atoms.