EDIT: Michael's upper bound is locally optimal, not globally as I had originally stated. Specifically, there is some neighborhood $N$ of the distribution $\alpha:=\frac{n}{n+1}\delta_0+\frac1{n+1}\delta_{n+1}$ (in the weak topology) such that $\mathbb P(X_1+\ldots+X_n<n+1)<\mathbb P(Y_1+\ldots+Y_n<n+1)$ whenever $(X_i),(Y_i)$ are iid with $X_1\sim\alpha$ and $Y_1\sim\mu\in N\setminus\{\alpha\}$.
To see this, let $S$ be the set of probability measures $\mu$ on $[0,\infty)$ such that $\int_0^\infty x\,\mu(dx)=1$, and for $\mu\in S$ define
$$E_n[\mu]:=\int_{\sum_{i=1}^nx_i<n+1}\mu(dx_1)\ldots\mu(dx_n).$$
Note that if $(X_i)$ are iid nonnegative random variables with $\mathbb E[X_1]=1$, and $\mu$ is the law of $X_1$, then $E_n[\mu]=\mathbb P(X_1+\ldots+X_n<n+1)$. Let $\alpha=\frac{n}{n+1}\delta_0+\frac1{n+1}\delta_{n+1}$, i.e. $\alpha$ is the distribution of the random variable Michael defines for the upper bound.
Observe that $S$ is convex, so given arbitary $\mu\in S\setminus\{\alpha\}$ the function $\Phi_n(t):=E_n[(1-t)\alpha+t\mu]$ is well-defined for $t\in[0,1]$. The formula for $\Phi_n(t)$ is complicated, but we do not need much of it:
$$\Phi_n(t)=c_0+\left(\sum_{i=1}^n\int_{\sum_jx_j<n+1}\mu(dx_i)\prod_{k\neq i}\alpha(dx_k)-\int_{\sum_jx_j<n+1}n\prod_{k=1}^n\alpha(dx_k)\right)t+\sum_{i=2}^nc_it^i$$
where the $c_i$ are constants depending on $\mu$ but not $t$. This yields
\begin{align*}
\Phi_n'(0) &=n\left(\int_{\sum_j x_j<n+1}\mu(dx_1)\prod_{k=2}^n\alpha(dx_k)-\int_{\sum_j x_j<n+1}\prod_{k=1}^n\alpha(dx_k)\right)\\
&=n\left[\mathbb P\left(Y_1+\sum_{i=2}^nX_i<n+1\right)-\mathbb P\left(\sum_{i=1}^nX_i<n+1\right)\right]
\end{align*}
where $Y_1$ is a random variable with law $\mu$, independent of the iid random variables $X_i$ which have law $\alpha$. Note that since $\mu\neq\alpha$ and $Y_1\ge0$, we have $\mathbb P(Y_1<n+1)>1-\frac1{n+1}$ and hence
$$\mathbb P\left(Y_1+\sum_{i=2}^nX_i<n+1\right)=\mathbb P(Y_1<n+1)\mathbb P(X_1=0)^{n-1}>\left(1-\frac1{n+1}\right)^n$$
and thus $\Phi_n'(0)>0$. This implies that there exists $\delta>0$ such that $\Phi_n(0)<\Phi_n(t)$ for all $t\in(0,\delta)$; since $\mu$ was arbitrary, it follows that $E_n$ has a strict local minimum at $\alpha$.
EDIT: I had originally written here that $E_n$ is convex and conclude the result. However, as Michael points out in the comments, this assertion is not true. It may still be possible to continue with this argument and conclude that $\alpha$ is the global minimizer of $E_n$, but at this point the best I have is that $\alpha$ is a local minimizer.
After having seen fedja's comment and reading related material, I find it unlikely anyone will be able to solve it the generalized version here as it is an open research problem.
