I know that not every local seminormal ring is CohenMacaulay. But are 1dimensional local seminormal rings CohenMacaulay?

Your two sentences present some discrepancy. Not all 1dimensional local rings are seminormal, so the first sentence has nothing to do with the second. If all 1dimensional local rings were CohenMacaulay, then all local rings would be CohenMacaulay by virtue of the definition of the CohenMacaulay property. Seminormalization can make things worse. Three lines meeting in a point that are not contained in a plane (say the 3 coordinate axes in 3space) is seminormal and it is not Gorenstein. On the other hand, 3 lines meeting in a point and contained in a plane is Gorenstein, but it is not seminormal. Its seminormalization is exactly the above union of 3 lines meeting in a point that are not contained in a plane. 


I agree with Sándor's, I'm not quite sure what you mean. And JSpecter provides a counterexample. However, 1 dimensional seminormal rings are CohenMacaulay. Indeed, 1 dimensional reduced rings are CohenMacaulay (you just need a single nonzerodivisor). 

