Let $(Y_1,\dots, Y_k)$ be a random $k$-tuple uniformly distributed in $(\mathbb R/\mathbb Z)^k$.

Let $X_i = \sum_{j=1}^k i^j Y_j \in \mathbb R/\mathbb Z$.

Then the $X_i$ are $k$-wise independent (and uniformly distributed on $\mathbb R/\mathbb Z$), since the linear map sending $(Y_1,\dots, Y_k)$ to $(X_{i_1},\dots, X_{i_k})$ is given by a $k\times k$ integer matrix with nonzero (Vandermonde) determinant, and such maps preserve the uniform measure on the torus $(\mathbb R/\mathbb Z)^k$.

(One way to check this is by Weyl's criterion, that $X_{i_1},\dots, X_{i_k}$ are uniformly distributed if and only if the expectation of $e^{ 2 \pi i \sum_{j=1}^k X_{i_j} n_j}$ is $0$ for all tuples $n_1,\dots, n_k$ of integers other than $0,\dots, 0$. This follows from uniform distribution of the $Y_1,\dots, Y_k$ because $\sum_{j=1}^k X_{i_j} n_j = \sum_{\ell=1}^k (\sum_{j=1}^k n_j i_j^\ell) Y_\ell $ and $(\sum_{j=1}^k n_j i_j^\ell)$ are integers not all $0$ by the nonvanishing of the Vandermonde. Another way is to note that each possible value of $X_1,\dots, X_k$ has a number of preimages equal to the absolute value of the determinant which cancels the inverse factor of the determinant of the derivative, which is the inverse of the Vandermonde determinant, from the change-of-variables formula in integration.)

But we can write $ k! Y_k = \sum_{j=0}^{k-1} (-1)^j \binom{k-1}{j} X_{i-j}$ for any $i$ so $k! Y_k$ is a nonconstant function in the tail sigma-algebra.