period of the lfsr sequence after some manipulations
If you pick every 4th term in the sequence, since the original period $P$ is
$$
(2^{L_1}-1)(2^{L_2}-1)(2^{L_3}-1)=2^{L_1+L_2+L_3}-2^{L_1+L_2}-2^{L_1+L_3}-2^{L_2+L_3}+
\sum_{i=1}^3 2^{L_i}-1,$$
and relatively prime to 4, except for trivially small choices for the $L_i,$ you get a proper decimation and a reordering of the sequence. There is no truncation.

Thus $LC_{new} \leq LC_{old}.$ But note that each of the component maximal length sequences can be written as $\mathrm{tr}(c_i \alpha_i^t),$ where $\alpha_i$ is a primitive element in $GF(2^{L_i})$ and the constant $c_i$ from the same field is used to adjust the phase.

However, the trace map is linear under repeated squaring $$\mathrm{tr}(x)=\mathrm{tr}(x^2)=\mathrm{tr}(x^4),$$ (constant on cyclotomic cosets) and decimation by 4 is equivalent to taking the fourth power, so I contend that the linear complexity also stays the same.

In addition, the weight of the decimated sequence obviously doesn't change.

The weight of the original sequence will simply be $(P+1)/2,$ since the function is balanced and the all zero point $f(0,0,0)=0$ is missing from the truth table.