We can in principle find the optimal first $n$ moments for a polynomial approximation of $f$. This reduces the problem to a truncated Hausdorff moment problem, for which solutions exist.

Let $f \colon [0,1]^d \to \mathbb{R}$ be a continuous function. Approximate $|f|$ with the multivariate Bernstein polynomials of order $n$ (writing $f_{k_1\cdots k_d} = f(k_1/n, \ldots k_d/n)$ for short),

$$|f_n|(x) = \sum_{k_1 \cdots k_d} |f_{k_1\cdots k_d}| \prod_{i=1}^{d} \binom{n}{k_i} \, x_i^{k_i} (1-x_i)^{n-k_i} \;.$$

Let $S_k = E[X^k (1-X)^{n-k}]$. The expectation of the approximant is then

$$E[|f_n|(X)] = \sum_{k_1 \cdots k_d} |f_{k_1\cdots k_d}| \prod_{i=1}^{d} \binom{n}{k_i} \, S_{k_i} \;,$$

which should be maximal. The value of $S_{k}$ is a linear combination on the first $n$ moments $M$ of the resulting distribution. Expanding

$$S_k = \sum_{r=0}^{n} \binom{n-k}{r-k} (-1)^{r-k} \, M_r \;,$$

the expectation in terms of the moments is

$$E[|f_n|(X)] = \sum_{k_1 \cdots k_d} \sum_{r_1 \cdots r_d} |f_{k_1\cdots k_d}| \prod_{i=1}^{d} \binom{n}{r_i} \binom{r_i}{k_i} (-1)^{r_i-k_i} \, M_{r_i} \;.$$

The problem is hence the constrained multilinear optimisation

$$
\begin{aligned}
\max_{M_0 \cdots M_n} \;&\; \sum_{k_1 \cdots k_d} \sum_{r_1 \cdots r_d} |f_{k_1\cdots k_d}| \prod_{i=1}^{d} \binom{n}{r_i} \binom{r_i}{k_i} (-1)^{r_i-k_i} \, M_{r_i} \;, \\
\text{s.t.} \;&\; \sum_{r=0}^{n} \binom{n-k}{r-k} (-1)^{r-k} \, M_r \ge 0 \;, \\
&\; M_r \in [0,1] \;, \quad M_0 = 1 \;.
\end{aligned}
$$

To make the problem slightly more tractable, the relation between $S_k$ and $M_r$ can be inverted using the binomial partial sums,

$$M_r = \sum_{k=0}^{n} \binom{n-r}{k-r} \, S_k \;,$$

so that the optimisation becomes

$$
\begin{aligned}
\max_{S_0 \cdots S_n} \;&\; \sum_{k_1 \cdots k_d} |f_{k_1\cdots k_d}| \prod_{i=1}^{d} \binom{n}{k_i} \, S_{k_i} \;, \\
\text{s.t.} \;&\; \sum_{k=0}^{n} \binom{n-r}{k-r} \, S_k \in [0,1] \;, \\
&\; \sum_{k=0}^{n} \binom{n}{k} \, S_k = 1 \;, \quad S_k \ge 0 \;.
\end{aligned}
$$

However, the first constraint is actually superfluous, as it is implied by the normalisation. Letting further $p_k = \binom{n}{k} S_k$, we have the optimisation problem in its most basic form,

$$
\begin{aligned}
\max_{p_0 \cdots p_n} \;&\; \sum_{k_1 \cdots k_d} |f_{k_1\cdots k_d}| \prod_{i=1}^{d} p_{k_i} \;, \\
\text{s.t.} \;&\; \sum_{k=0}^{n} p_k = 1 \;, \quad p_k \ge 0 \;.
\end{aligned}
$$

which is simply the maximum expectation of the discretised function $|f_{k_1 \cdots k_d}|$ from the discrete distribution $p$.

From a cursory Google search, multilinear optimisation appears difficult (but I have no idea, really). Nevertheless, it is a recipe for calculations: For the baby problem, I find (with $n = 50$) the expectation 0.0615760 and distribution $p$ as below.

[![discrete distribution][1]][1]

  [1]: https://i.sstatic.net/25rMK.png