Your space contains an isomorphic copy of $L^\infty([0,1])$ (consider the family of elements with support in the interval) and so is not nuclear.
By the way an explanation of the relationship between smoothness and nuclearity is that many spaces of test functions are generated in a natural way by differential operators as described in a short but important article by A. Pietsch (Math. Ann. vol. 164). This article contains a simple criterion for such spaces to be nuclear, based on the spectral properties of the operators as unbounded, self-adjoint ones on Hilbert space and displays some of the standard spaces as explicit examples.