Suppose $L$ is a nilpotent finite-dimensional Lie algebra over $\mathbb{Q}$ of class $c$. We can define an associated graded Lie algebra to $L$ that, as a vector space, is:

$$\bigoplus_{i=1}^c \gamma_i(L)/\gamma_{i+1}(L)$$

where $\gamma_i$ denotes the $i^{th}$ member of the lower central series, and where the Lie bracket is obtained from the Lie bracket of $L$ in a natural fashion (as follows: the Lie bracket of $L$ induces a map $\gamma_i(L) \times \gamma_j(L) \to \gamma_{i+j}(L)$ which descends to a bilinear map $\gamma_i(L)/\gamma_{i+1}(L) \times \gamma_j(L)/\gamma_{j+1}(L) \to \gamma_{i+j}(L)/\gamma_{i+j+1}(L)$. The Lie bracket in the associated graded is obtained from these constituent maps being summed up.

I believe that, in general, $L$ and its associated graded Lie algebra do not necessarily have to be isomorphic. However, I'm having a little trouble coming up with an explicit example where they're not. In particular: (i) if $L$ has class two, then it is isomorphic to its graded Lie algebra, (ii) if $L$ is a "free nilpotent" Lie algebra on a certain number of generators, then again it is isomorphic to its graded Lie algebra.

This has some relation with the Malcev Lie correspondence, but I preferred not to introduce the complication of nilpotent groups into the question. If viewed that way, the question is: for a rationally powered finitely generated nilpotent group, is the Malcev Lie ring the same as its associated graded Lie ring? Some of the discussion at Relationship between the cohomology of a group and the cohomology of its associated Lie algebra. may also be relevant.