Here is a solution in the special case in which each lower central factor $\gamma_i(G) / \gamma_{i+1}(G)$ is elementary abelian.
First consider the case in which $G$ is elementary abelian, written additively, so $G \cong \mathbf F_p^d$ for some $d$. Let $x$ be a generator of $C_p$. Then for $g \in G^p$ we have $[g,x] = \Delta(g)$, where $\Delta = \sigma-1$ and $\sigma : G^p \to G^p$ is the linear map defined by $\sigma(g_1, \dots, g_p) = (g_p, g_1, \dots, g_{p-1})$. Therefore
$$\gamma_i(G \wr C_p) = \Delta^i(G^p) \qquad (i \ge 2).$$
By the $\mathbf{F}_p[X]$ polynomial identity
$(X-1)^{p-1} = 1 + X + \cdots + X^{p-1}$
we have
$$\Delta^{p-1} = 1 + \sigma + \cdots + \sigma^{p-1}.$$
Therefore
$$\Delta^{p-1}(g) = (s, \dots, s), \qquad \text{where}~s=g_1 + \cdots + g_p.$$
Therefore $\gamma_{p-1}(G \wr C_p) = \Delta^{p-1}(G^p)$ is just the diagonal copy of $G$ in $G^p$, which is also the kernel of $\Delta$. It follows that the lower central factors of $G \wr C_p$ are $G \times C_p, G, \dots, G$.
Now consider the general case. Let $G = \Gamma_1 \ge \cdots \ge \Gamma_k = 1$ be the lower central series of $G$. Then $C_p$ acts on each factor $\Gamma_i^p / \Gamma_{i+1}^p$, and the lower central series of $\Gamma_i / \Gamma_{i+1} \wr C_p$ is as described above. By lifting to $G$ we get a long central series:
$$G \wr C_p = \Gamma_1^p C_p \ge \Delta(\Gamma_1^p) \Gamma_2^p \ge \Delta^2(\Gamma_1^p) \Gamma_2^p \ge \cdots \ge \Delta^{p-1}(\Gamma_1^p) \Gamma_2^p \ge \Gamma_2^p \ge \Delta(\Gamma_2^p) \Gamma_3^p \ge \cdots.$$
I claim this is in fact the lower central series. To see this it suffices to observe that $\Delta^i(\Gamma_j^p)\Gamma_{j+1}^p$ (for $i \le p-1$) contains the preimage $D$ in $\Gamma_j^p$ of the diagonal subgroup of $\Gamma_j^p/\Gamma_{j+1}^p$, and $[D, G^p] = \Gamma_{j+1}^p$.
In general if $\gamma_i(G)/\gamma_{i+1}(G)$ is not elementary abelian it seems more complicated. For example $C_{p^2} \wr C_p$ seems to have lower central factors $C_{p^2} \times C_p, C_p, C_p, \dots, C_p$.