Disclaimer: This is a bit too much "abstract nonsense" for me to really consider it a satisfying answer to the question, but on the other hand the question is so general that I'm not sure how much better one can do.
Let's write $E=[0,1]$ and $P(E)$ for probability measures on $E$. Let $f : E^n \to \mathbb{R}$ be bounded and measurable. (Actually, this all holds for an arbitrary Polish space $E$.) I claim that you can write
$$\sup_{\mu \in P(E)}\int_{E^n} f\,d\mu^n = \lim_{k\to\infty}\sup_{(x_1,\ldots,x_k) \in E^k}\frac{1}{k!}\sum_{\sigma \in S_k}f(x_{\sigma(1)},\ldots,x_{\sigma(n)}),$$
where $S_k$ is the set of permutations of $\{1,\ldots,k\}$, and $\mu^n$ is the $n$-fold product measure. As a sanity check, note that if $n=1$ then you can take $x_1=\ldots=x_k$ in the supremum to get $\sup f$. Not sure what to do with this for $n\ge 2$.
Proof: Fix $k \ge n$ for now. Then
$$\int_{E^n}f\,d\mu^n = \int_{E^k}f(x_1,\ldots,x_n)\,\mu^k(d(x_1,\ldots,x_k))$$
is a linear functional of $\mu^k$. The supremum of a linear functional over a convex set is the same as over its closed convex hull. Thus, if $C_k \subset P(E^k)$ denotes the closed (in total variation) convex hull of $\{\mu^k : \mu \in P(E)\}$, then
$$\sup_{\mu \in P(E)}\int_{E^n}f\,d\mu^n = \sup_{\nu \in C_k}\int_{E^k}f(x_1,\ldots,x_n)\,\nu(d(x_1,\ldots,x_k)).$$
Note that $C_k$ consists of all "mixtures of iid". Because De Finetti's theorem fails for finite exchangeable sequences, these "mixtures of iid" are not quite all exchangeable probability measures on $E^k$. However, if $P_e(E^k)$ denotes the set of exchangeable (invariant under permutations) probability measures on $E^k$, then a theorem of Diaconis-Freedman (Theorem 13 here) says that for each $\eta \in P_e(E^k)$ there exists $\nu \in C_k$ such that the projections $\eta|_n$ and $\nu|_n$ onto the first $n$ coordinates satisfy $\|\eta|_n-\nu|_n\|_{TV} \le n(n-1)/2k$, with $\|\cdot\|_{TV}$ denoting total variation. It follows that
$$\left|\sup_{\mu \in P(E)}\int_{E^n}f\,d\mu^n - \sup_{\eta \in P_e(E^k)}\int_{E^k}f(x_1,\ldots,x_n)\,\eta(d(x_1,\ldots,x_k))\right| \le \|f\|_\infty n(n-1)/2k.$$
Next, for a given $\eta$ we can average over all permutations of coordinates in $f$ without changing the value of the integral. That is,
$$\sup_{\eta \in P_e(E^k)}\int_{E^k}f(x_1,\ldots,x_n)\,\eta(d(x_1,\ldots,x_k)) = \sup_{\eta \in P_e(E^k)}\int_{E^k}\hat{f}_k\,d\eta,$$
where the function $\hat{f}_k : E^k \to \mathbb{R}$ is defined by
$$\hat{f}_k(x_1,\ldots,x_k) = \frac{1}{k!}\sum_{\sigma \in S_k}f(x_{\sigma(1)},\ldots,x_{\sigma(n)}).$$
We can then "un-symmetrize" the measures $\eta \in P_e(E^k)$ to reach a supremum over all probabilities on $E^k$:
$$\sup_{\eta \in P_e(E^k)}\int_{E^k}\hat{f}_k\,d\eta = \sup_{\eta \in P(E^k)}\int_{E^k}\hat{f}_k\,d\eta.$$
The right-hand side is simply $\sup \hat{f}_k$. Putting it all together gives
$$\sup_{\mu \in P(E)}\int_{E^n}f\,d\mu^n = \lim_{k\to\infty}\sup_{\eta \in P_e(E^k)}\int_{E^k}f(x_1,\ldots,x_n)\,\eta(d(x_1,\ldots,x_k)) = \lim_{k\to\infty}\sup \hat{f}_k,$$
which was the original claim above.