Given the Jacobi identity $[\mathbb{X},[\mathbb{Y},\mathbb{Z}]]+[\mathbb{Y},[\mathbb{Z},\mathbb{X}]]+[\mathbb{Z},[\mathbb{X},\mathbb{Y}]]=0$ and that the Lie derivative of a vector field is $L_{\mathbb{X}}\mathbb{Y}=[\mathbb{Y},\mathbb{X}]$.
One can show that the Jacobi identity can also be written as $L_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}]=[L_{\mathbb{X}}\mathbb{Y},\mathbb{Z}]+[\mathbb{Y},L_{\mathbb{X}}\mathbb{Z}]$
Then by using induction show that $\displaystyle L^n_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}]=\sum_{j=0}^{n}\binom nj [L^j_{\mathbb{X}}\mathbb{Y} ,L^{n-j}_{\mathbb{X}}\mathbb{Z}]$
Hint use the identity $\binom n{j-1}+\binom n{j}=\binom {n+1}{j}$
The case $n=1$ is clearly true. I am stuck with the inductive step, assuming true for $n=k$,
$\begin{align} L^{k+1}_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}] &= L_{\mathbb{X}} (L^{k}_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}]) \\ &= L_{\mathbb{X}}( \sum_{j=0}^{k}\binom kj [L^j_{\mathbb{X}}\mathbb{Y} ,L^{k-j}_{\mathbb{X}}\mathbb{Z}]) \\ &= \sum_{j=0}^{k}\binom kj L_{\mathbb{X}}( [L^j_{\mathbb{X}}\mathbb{Y} ,L^{k-j}_{\mathbb{X}}\mathbb{Z}]) \\ &= \sum_{j=0}^{k}\binom kj ([L_{\mathbb{X}}(L^j_{\mathbb{X}}\mathbb{Y}) ,L^{k-j}_{\mathbb{X}}\mathbb{Z}]+[L^j_{\mathbb{X}}\mathbb{Y} ,L^j_{\mathbb{X}}(L^{k-j}_{\mathbb{X}}\mathbb{Z})]) \\ &= \sum_{j=0}^{k}\binom kj ([L^{j+1}_{\mathbb{X}}\mathbb{Y} ,L^{k-j}_{\mathbb{X}}\mathbb{Z}]+[L^j_{\mathbb{X}}\mathbb{Y} , L^{k-j+1}_{\mathbb{X}}\mathbb{Z}]) \end{align}$
and then I get stuck
For $n=1$ it is clear (as you said, $L_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}] = [L_{\mathbb{X}}\mathbb{Y},\mathbb{Z}] + [\mathbb{Y},L_{\mathbb{X}}\mathbb{Z}]$).
Assume it is true for $n-1$, and let's prove the case $n$:
\begin{equation*} L^n_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}] =L_{\mathbb{X}} ( L^{n-1}_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}]) = L_{\mathbb{X}}\Big( \sum_{j=0}^{n-1} \binom {n-1}j [L_{\mathbb{X}}^j \mathbb{Y} , L_{\mathbb{X}}^{n-1-j} \mathbb{Z}] \Big) = \sum_{j=0}^{n-1} \binom {n-1}j L_{\mathbb{X}} \Big( [L_{\mathbb{X}}^j \mathbb{Y} , L_{\mathbb{X}}^{n-1-j} \mathbb{Z}]\Big) \end{equation*}
Now use the case $n=1$ to get
\begin{equation*} L_{\mathbb{X}} \Big( [L_{\mathbb{X}}^j \mathbb{Y} , L_{\mathbb{X}}^{n-1-j} \mathbb{Z}]\Big) = [L_{\mathbb{X}}^{j+1} \mathbb{Y} , L_{\mathbb{X}}^{n-1-j} \mathbb{Z}] + [L_{\mathbb{X}}^j \mathbb{Y} , L_{\mathbb{X}}^{n-j} \mathbb{Z}] \end{equation*}
Now, you get
\begin{equation*} L^n_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}] = \sum_{j=0}^{n-1} \binom {n-1}j [L_{\mathbb{X}}^{j+1} \mathbb{Y} , L_{\mathbb{X}}^{n-1-j} \mathbb{Z}] + \sum_{j=0}^{n-1} \binom {n-1}j [L_{\mathbb{X}}^j \mathbb{Y} , L_{\mathbb{X}}^{n-j} \mathbb{Z}] = \sum_{j=1}^{n} \binom {n-1}{j-1} [L_{\mathbb{X}}^{j} \mathbb{Y} , L_{\mathbb{X}}^{n-j} \mathbb{Z}] + \sum_{j=0}^{n-1} \binom {n-1}j [L_{\mathbb{X}}^j \mathbb{Y} , L_{\mathbb{X}}^{n-j} \mathbb{Z}] = \end{equation*}
\begin{equation*} = [L_{\mathbb{X}}^n \mathbb{Y},L_{\mathbb{X}}^0\mathbb{Z}] + \sum_{j=1}^{n-1} \binom {n-1}{j-1} [L_{\mathbb{X}}^{j} \mathbb{Y} , L_{\mathbb{X}}^{n-j} \mathbb{Z}] + \sum_{j=1}^{n-1} \binom {n-1}j [L_{\mathbb{X}}^j \mathbb{Y} , L_{\mathbb{X}}^{n-j} \mathbb{Z}] + [L_{\mathbb{X}}^0 \mathbb{Y} , L_{\mathbb{X}}^{n} \mathbb{Z}] \end{equation*}
Now rearranging and using the hint,
\begin{equation*} L^n_{\mathbb{X}}[\mathbb{Y},\mathbb{Z}] = [L_{\mathbb{X}}^n \mathbb{Y},L_{\mathbb{X}}^0\mathbb{Z}] + \sum_{j=1}^{n-1} \Big[ \binom {n-1}{j-1} + \binom {n-1}j \Big] [L_{\mathbb{X}}^j \mathbb{Y} , L_{\mathbb{X}}^{n-j} \mathbb{Z}] + [L_{\mathbb{X}}^0 \mathbb{Y} , L_{\mathbb{X}}^{n} \mathbb{Z}] = [L_{\mathbb{X}}^n \mathbb{Y},L_{\mathbb{X}}^0\mathbb{Z}] + \sum_{j=1}^{n-1} \binom {n}{j} [L_{\mathbb{X}}^j \mathbb{Y} , L_{\mathbb{X}}^{n-j} \mathbb{Z}] + [L_{\mathbb{X}}^0 \mathbb{Y} , L_{\mathbb{X}}^{n} \mathbb{Z}] \end{equation*}
\begin{equation*} = \sum_{j=0}^{n} \binom {n}{j} [L_{\mathbb{X}}^j \mathbb{Y} , L_{\mathbb{X}}^{n-j} \mathbb{Z}] \end{equation*}
as required.