Following NN2's idea, I figured out a generalization for the formula. If we note:
$$\sum\limits_{n=1}^{\infty}T_{n}(x) \frac{t^n}{n} =
- \frac{1}{2} \,
\ln \bigg( 1 + P_0(x)\,t + t^2 \bigg) \ \ \ \ with \ \ \ P_0(x)=-2x
$$
then we will have:
$$\sum\limits_{n=1}^{\infty}T_{2n}(x) \frac{t^n}{n} =
- \frac{1}{2} \,
\ln \bigg( 1 + P_1(x)\,t + t^2 \bigg)
$$
with
$$ P_1(x)=2-P_0^2(x)$$
More generally, we will have :
$$\sum\limits_{n=1}^{\infty}T_{n\,2^k}(x) \frac{t^n}{n} =
- \frac{1}{2} \,
\ln \bigg( 1 + P_k(x)\,t + t^2 \bigg) \ \ \ \ with \ \ \ P_{k}(x) = 2 - P_{k-1}^2(x)
$$
The first $P_n$ polynomials are:
\begin{equation}
\begin{split}
P_0(x) &= -2x \\
P_1(x) &= -4x^2 + 2 \\
P_2(x) &= -16x^4 + 16x^2 - 2 \\
P_3(x) &= -256x^8 + 512x^6 - 320x^4 + 64x^2 - 2 \\
\end{split}
\end{equation}