I wanted to show that for $A = K[X_i, i \in \mathbb{N}]/(X_iX_j)_{i,j \in \mathbb{N}}$ the completion $A[T] \rightarrow A[|T|]$ is not flat.
However, my proof seems a bit simple/ direct to me, so I'm a little worried that I made a stupid flaw.
Consider the inclusion of the principal ideal $(X_0) \rightarrow A$.
Then after tensoring with $A[|T|]$ over $A$, it is $X_0T \neq 0 \in (X_0) \otimes A[|T|]$, but in $A[|T|]$: $X_0T= X_0(T+ \sum_{i \geq 2} X_iT^i) = X_0T(1+ \sum_{i \geq 2} X_i T^{i-1})$.
The right factor is a unit, hence $X_0T=0$ or $\sum_{i \geq 2} X_i T^{i-1} = 0$. In a similar manner I can show $\sum_{i \geq 3} X_i T^{i-1} =0$ and deduce that $X_iT=0$ for all $i$.
Aside from a correction/pointing out the mistake in my proof, I'd be very happy to see more examples of non-flat completion.
Although I don't have an elementary argument for your particular case, let me say that any ring which is not coherent is good (see here), and it's easy to check that your ring isn't coherent.
Other example: $A=K[Y,X_0,\dots,X_n,\dots]/(YX_0, X_0-YX_1,\dots,X_n-YX_{n+1},\dots)$. Note that $y-T$ is a non-zero divisor in $A[T]$ and $(y-T)\sum_{i\ge 0} x_iT^i=0$ in $A[[T]]$.