This was a funny problem. The answer is $\lambda \ge e$. We first show that all $\lambda < e$ fail. Rewrite the inequality (multiply both sides by $a_1a_2\ldots a_n$) as $\sum{a_1a_2\ldots a_{i-1}a_{i+1}\ldots a_n}-\lambda \le (n-\lambda)\cdot a_1a_2\ldots a_n \text{ }(*)$. Now set $a_n = 0, a_1=a_2=\ldots=a_{n-1} = \frac{n}{n-1}$. Our inequality reduces to $\left(\frac{n}{n-1}\right)^{n-1} - \lambda \le 0 \implies \lambda \ge \left(\frac{n}{n-1}\right)^{n-1}$. Taking the limit of both sides as $n \rightarrow \infty$, we get $\lambda \ge e$. Now we show that all $\lambda \ge e$ in fact satisfy the inequality. For this Lagrange Murderpliers work really nicely. By LM, the LHS of the original inequality is maximized when the following things hold: $-\frac{1}{a_i^2}+\frac{\lambda}{a_i}\cdot\frac{1}{a_1a_2\ldots a_n} = \mu \implies$ \[\frac{\lambda}{a_1a_2\ldots a_n} = \mu a_i +\frac{1}{a_i} \text{ for } i = 1, 2, \ldots, n.\] So assume that $a_i \not= a_j$ for some $i, j$. Then $\mu a_i +\frac{1}{a_i} = \mu a_j +\frac{1}{a_j} \implies \mu = \frac{1}{a_ia_j}$ after easy algebra. Plugging this in for $\mu$, and using AM-GM, we get $\frac{\lambda}{a_1a_2\ldots a_n} = \frac{1}{a_j} +\frac{1}{a_i} \implies \lambda = \frac{a_i+a_j}{a_ia_j} \cdot a_1a_2\ldots a_n \le \left(\frac{a_1+a_2+\ldots+a_n}{n-1}\right)^{n-1} = \left(\frac{n}{n-1}\right)^{n-1} < e$, contradiction. So $a_i = a_j$. So all the things are equal! The inequality is easy to verify in this case. Now we only need to check boundary cases. WLOG, set $a_n = 0$ and look at the $(*)$ form of the inequality. It suffices to show that $a_1a_2\ldots a_{n-1} \le \lambda$. But once again, by AM-GM, $ a_1a_2\ldots a_{n-1} \le \left(\frac{n}{n-1}\right)^{n-1} \le e \le \lambda. _\blacksquare$

We prove that $\lambda \geq e$ satisfies (the rest is same as above)$.$ Let $f(a_1,a_2,\cdots ,a_n)=\sum_{i=1}^n\frac{1}{a_i}-\lambda\prod_{i=1}^{n}\frac{1}{a_i}.$ Lemma $f(a_1,a_2,\cdots ,a_n)\leqslant f(1,a_1+a_2-1,\cdots ,a_n).$ $\Leftrightarrow\sum_{i=1}^n\frac{1}{a_i}-\lambda\prod_{i=1}^{n}\frac{1}{a_i}\leqslant 1+\frac{1}{a_1+a_2-1}+\sum_{i=3}^n\frac{1}{a_i}-\frac{\lambda}{a_1+a_2-1}\prod_{i=3}^{n}\frac{1}{a_i}.$ $\Leftrightarrow\frac{a_1+a_2}{a_1a_2}-\frac{a_1+a_2}{a_1+a_2-1}\leqslant\lambda\left(\frac{1}{a_1a_2}-\frac{1}{a_1+a_2-1}\right)\prod_{i=3}^{n}\frac{1}{a_i}\Leftrightarrow (a_1+a_2)a_3a_4\cdots a_n\leqslant \lambda.$ Using AM-GM inequality$,$ we have $(a_1+a_2)a_3a_4\cdots a_n\leqslant\left(\frac{n}{n-1}\right)^{n-1}<e\leqslant\lambda.$ Solution $f(a_1,a_2,\cdots ,a_n)\leqslant f(1,a_1+a_2-1,\cdots ,a_n)\leqslant\cdots\leqslant f(1,1,\cdots ,1)=n-\lambda.\blacksquare$