$L_3 = L_1(L_2)^*$
So, We can write $L_3$ in the following way :
$L_3 = L_1(L_2^0 + L_2^1 + L_2^2 + ...)$
$L_3 = L_1(\left \{ \in \right \} + L_2 + L_2.L_2 + L_2.L_2.L_2 + ...)$
$L_3 = ( L_1 + L_1.L_2 + L_1.L_2.L_2 + L_1.L_2.L_2.L_2 + ...)$
Now, Because in $L_1$, $n$ is greater than $1$. $L_1$ can't have null string and which makes all the strings of the language $L_3$ starting with $a^nb^nc^n$ where $n \geq 1$ Which makes $L_3$ CSL because whatever logic we apply for $L_3$, We have to parse $a^nb^nc^n$ before proceeding further in the string.
Hence, $L_3$ is $CSL$.
Note that if $L_1$ had contained $\in$ i.e. if $n \geq 0$ then $L_3$ will be Regular. Moreover, in that case $L_3$ would be $\Sigma^*$.