Here is a succinct description of $f_{\epsilon}(n,d)$ (I only have time to sketch the argument-- hopefully it is right).
Let
$$
h(x_1,\dots, x_d)=\left(\sum_{1 \leq i_1 < \dots < i_{n+1} \leq d} x_{i_1}\cdots x_{i_{n+1}}\right),
$$
and let $x \in \mathbb{R}^d$ be a feasible element of the minimization (i.e., $x_1\geq \dots \geq x_d\geq 0, x_1+\dots+x_n=1-\epsilon$, and $x_{n+1}+\dots+x_d=\epsilon$).
For any $i<j \in \{n+1,\dots, d\}$, let
$$g(t)=h(x_1,\dots, x_{i-1},x_i+t,x_{i+1},\dots,x_{j-1},x_j-t, x_{j+1},\dots, x_d).$$
It is straightforward to verify that $g(t)$ is a quadratic function with negative second derivative. So $g(t)$ is minimized at an extreme point. It follows that $h(x)$ is minimized when $x_{r+1},\dots, x_n$ are as ``unbalanced" as possible. This implies that the minimum is attained for some $x \in \mathbb{R}^d$ for which
$$
(x_{n+1},\dots, x_d)=\left(\frac{1-\epsilon}{n},\dots, \frac{1-\epsilon}{n},\epsilon-\frac{k(1-\epsilon)}{n},0,\dots,0\right),
$$
where $\frac{1-\epsilon}{n}$ is repeated $k$ times in the tuple, and
$$
k=\lfloor \frac{\epsilon n}{1-\epsilon} \rfloor.
$$
We can do the same trick for the first $n$ coordinates of $x$, and prove that the minimum is attained for $x \in \mathbb{R}^d$ that satisfies
$$
(x_1,\dots, x_n)=\begin{cases} (\frac{1-\epsilon}{n},\dots,\frac{1-\epsilon}{n}) ,& \epsilon \geq \frac{1-\epsilon}{n}\\
(1-n\epsilon,\epsilon,\dots, \epsilon) ,& \epsilon\leq \frac{1-\epsilon}{n}.
\end{cases}
$$
It follows that
$$
f_\epsilon(n,d)=\begin{cases} \binom{n+k}{n+1} \left(\frac{1-\epsilon}{n}\right)^{n+1}+\binom{n+k}{n} \left(\frac{1-\epsilon}{n}\right)^n \left[\epsilon-k\left(\frac{1-\epsilon}{n}\right)\right] ,& \epsilon \geq \frac{1-\epsilon}{n}\\
(1-n\epsilon)\epsilon^n,& \epsilon\leq \frac{1-\epsilon}{n}.
\end{cases}
$$
