There's probably no "closed form" for any $s>1$. Curiously, though,
the sum can be evaluated for integers $s \leq 0$, in the following sense.
The function $\zeta_{TM}$ extends to an analytic function on
${\bf C} \backslash \{ 1 \}$, with a simple pole at $s=1$
of residue $1/2$, and taking rational values at integers $s \leq 0$,
starting
$\zeta_{TM}(0) = -1/4$,
$\zeta_{TM}(-1) = -1/24$,
$\zeta_{TM}(-2) = 0$,
$\zeta_{TM}(-3) = +1/240$,
and in general
$\zeta_{TM}(s) = \zeta(s) / 2$ for integers $s \leq 0$
(so in particular $\zeta_{TM}$ inherits the "trivial zeros" of $\zeta$ at
$s = -2, -4, -6, \ldots$).
It is more convenient to work with the Dirichlet series whose
$(n+1)^{-s}$ coefficient is $1 - 2 t_n = (-1)^{t_n}$, because
the generating function for $(-1)^{t_n}$, call it
$$
T(z) = \sum_{n=0}^\infty (-1)^{t_n} z^n,
$$
factors as an infinite product:
$$
T(z) = (1-z) (1-z^2) (1-z^4) (1-z^8) \cdots
= \prod_{m=0}^\infty \bigl( 1 - z^{2^m} \bigr).
$$
So define
$$
Z_{TM}(s) = \zeta(s) - 2 \zeta_{TM}(s)
= \sum_{n=0}^\infty \frac{(-1)^{t_n}}{(n+1)^s}.
$$
The usual Mellin-transform trick gives an integral formula:
$$
\Gamma(s) Z_{TM}(s)
= \sum_{n=0}^\infty (-1)^{t_n} \! \int_0^\infty x^{s-1} e^{-(n+1)x} \, dx
= \int_0^\infty x^{s-1} e^{-x} T(e^{-x}) \, dx.
$$
This gives an analytic continuation of $\Gamma(s) Z_{TM}(s)$ to the entire
complex plane, because $T(e^{-x})$ decays faster than any power of $x$
as $x \to 0$: each factor $1 - e^{-2^m x}$ of the infinite product
is $O_m(x)$ and in $(0,1)$. Since $\Gamma(s)$ has no zeros,
but does have simple poles at $s = 0, -1, -2, -3, \ldots$,
it follows that $Z_{TM}$ is an entire function with simple zeros
at the same $s$, and no other real zeros (the integral for
$\Gamma(s) Z_{TM}(s)$ is plainly positive for all real $s$).
Since $Z_{TM} = \zeta - 2 \zeta_{TM}$, we conclude that
$\zeta_{TM}(s) = \frac12 \zeta(s)$ at those $s$, as claimed.
[The integral formula can also be used to compute
$Z_{TM}(s)$, and thus also $\zeta_{TM}(s)$, to high precision;
for example, using gp's "intnum" function we find
$Z_{TM}(2) = 0.6931534522\ldots$ (this is not $\log 2$,
though it's quite close -- the difference is $\lt 10^{-5}$),
so $\zeta_{TM}(2) = (\zeta(2) - Z_{TM}(2)) / 2 = 0.4758903073\ldots$.]
Using the language of formal power series, this is moderately easy to check.
Let $A(z)=\sum_{n=1}^{\infty}a_n z^n$, considered as a formal power series over $\mathbb{R}$ (for a while). Multiplying $$na_n=(5-2n)a_{n-1}+3(n-1)a_{n-2}+1$$ by $z^{n-1}$ and summing (formally!) over $n>0$ (yes, $n=1$ included), we obtain $$A'(z)=\big(3A(z)-2zA'(z)\big)+3\big(zA(z)+z^2A'(z)\big)+(1-z)^{-1}.$$
Considered as an ODE in the regular sense, with $A(0)=0$, it is easily solved: $$A(z)=\frac{1}{2(1-z)}\left(\sqrt{\frac{1+3z}{1-z}}-1\right),$$ which has a power series expansion convergent for $|z|<1/3$. Being unique, it is exactly our $\sum_{n=1}^{\infty}a_n z^n$.
The function $A(z)$ satisfies $(1-z)^3 A^2(z)+(1-z)^2 A(z)=z$. As an equation for (again) a formal power series $A(z)=\sum_{n=1}^{\infty}a_n z^n$ over $\mathbb{R}$, it has a unique solution (this is seen from the recurrence relation obtained after substituting $A(z)=\sum_{n=1}^{\infty}a_n z^n$ into the equation, and a by-product is that $a_n$ are indeed integers). Now the important observation is that the same is true for a formal power series over the field $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ (for the same reason).
Thus, it just remains to check that the "generating function" of the Thue-Morse sequence $$B(z)=\sum_{n=0}^{\infty}b_n z^n,\qquad\color{blue}{b_0=0,}\ b_{2n}=b_n,\ b_{2n+1}=1-b_n,$$ considered as a formal power series over $\mathbb{F}_2$, satisfies the same equation. But $$B(z)=\sum_{n=0}^{\infty}b_{2n}z^{2n}+\sum_{n=0}^{\infty}b_{2n+1}z^{2n+1}=(1-z)B(z^2)+z(1-z^2)^{-1},$$ that is, $(1-z^2)B(z)-(1-z)(1-z^2)B(z^2)=z$; since $F(z)=-F(z)$ and $F(z^2)=F^2(z)$ for any formal power series $F(z)$ over $\mathbb{F}_2$, this gives $(1-z)^2 B(z)+(1-z)^3 B^2(z)=z$, as expected.
Best Answer
Given any integer $n \ge 0$, let $( n_0, n_1, n_2, \ldots )$ be its binary representation, i.e.
$$n = \sum_{i=0}^\infty n_i 2^i, \quad n_i \in \{ 0, 1 \}$$
Let $P(n) = n_0$ be the parity of $n$ and $N(n) = \sum\limits_{i=0}^\infty n_i$ be the number of set bits in this binary representation. It is not hard to see $t_n = 1$ when and only when $N(n)$ is odd. i.e. $$t_n = P(N(n))$$
Notice $$n - \sum_{k=1}^n (-1)^{\binom{n}{k}} = \sum_{k=0}^n \left(1 - (-1)^{\binom{n}{k}}\right) -2 = 2\sum_{k=0}^nP\left(\binom{n}{k}\right) - 2\tag{*1} $$ For any $0 \le k \le n$, let $(k_0,k_1,k_2,\ldots)$ be the binary representation of $k$.
By Lucas' theorem, we have $$P\left(\binom{n}{k}\right) = \prod_{i=0}^\infty P\left(\binom{n_i}{k_i}\right)$$
where $\displaystyle\;\binom{n_i}{k_i}$ should be interpreted as $0$ whenever $n_i < k_i$.
In order for the summand in RHS of $(*1)$ to be non-zero,
This means in the rightmost sum of $(*1)$, exactly $2^{N(n)}$ of $P(\cdot)$ contributes. This leads to
$$\begin{align} & n - \sum_{k=1}^n(-1)^{\binom{n}{k}} = 2^{N(n)+1} - 2 \equiv 2P(N(n)) \pmod 3\\ \implies & \frac43\sin^2\left(\frac{\pi}{3}\left(n - \sum_{k=1}^n (-1)^{\binom{n}{k}}\right)\right) = \frac43\sin^2\left(\frac{2\pi}{3}P(N(n))\right) \stackrel{\color{blue}{\because P(\cdot) = 0\text{ or } 1}}{=} P(N(n)) = t_n \end{align}$$