Yes. It's known to be transcendental. The sequence of coefficients of your number is a variant of a Sturmian sequence. It has very low complexity. The definition of this: let the digit sequence be $a_1,a_2,a_3\ldots$ taking values in $ \lbrace 0,1\ldots,d-1 \rbrace ^{\mathbb N}$. A subword of length $k$ is a string $a_ia_{i+1}a_{i+2}\ldots a_{i+k-1}$. The complexity, $p(k)$, is a function from $\mathbb N$ to $\mathbb N$ taking $k$ to the number of subwords of the sequence of length $k$.
In a 2007 paper in the Annals of Mathematics (vol 165, p547--565), Adamczewski and Bugeaud (On the complexity of algebraic numbers. I. Expansions in integer bases) showed that if a number is algebraic, then its digit sequence in base $b$ has complexity satisfying $p(k)/k\to\infty$.
In your case, the complexity of the sequence of base 2 digits satisfies $p(k)=2k$. How to see this?
Define a map $f$ from $[0,1)$ to $\lbrace 0,1\rbrace$ by $f(x)=1$ if $x\in [0,1/2)$ and 0 otherwise.
The $n$th term of your sequence is $f(\alpha n\bmod 1)$, where $\alpha=1/(2\pi)$. Write $T$ for the transformation from $[0,1)$ to itself given by $T(x)=x+\alpha\bmod 1$. Then the $n$th term is just $f(T^n0)$. The sub-block of the digit sequence of length $k$ starting at the $j$th term is $f(T^j0)\ldots f(T^{j+k-1}0)$. Since the $T^j0$ are dense in $[0,1)$, we need to ask how many blocks $f(x)\ldots f(T^{k-1}x)$ are possible.
Consider taking $x=0$ and moving it around the circle (=$[0,1)$) once. As you move it, the $T^ix$ also each move around the circle one time. The sequence changes each time one of the $(T^ix)_{0\le i< k}$ crosses 0 or 1/2. This is a total of $2k$ changes. Hence the sequence takes on $2k$ values as $x$ moves around the circle, hence the estimate for the complexity.
In his 1887 paper Table des valeurs des sommes $S_k = \sum_{1}^\infty n^{-k}$ (Acta Mathematica 10 (1887), 299-302; volume available online), Stieltjes used almost exactly this formula to compute Euler's constant to 33 decimal places. Of course as quid points out you need to know the zeta values to do this, but the main point of this paper was to compute those values, so he was just getting Euler's constant as a corollary. He uses a slight variant of the formula, with $\zeta(2k+1)-1$ in place of $\zeta(2k+1)$ for faster convergence (and a corresponding adjustment in the other term, which becomes $1+\log 2 - \log 3$). He derives the formula by taking the Taylor series expansion of $\log \Gamma(1+x)$ and using it to compute $\log \Gamma(1+1/2) - \log \Gamma(1-1/2)$.
Best Answer
There are number theorists who understand this subject much better than I do. However, I feel obliged to post an incomplete answer quickly before people have a chance to close this question.
There are a lot more connections known between $\pi$ and $e$ and other numbers than between $\gamma$ and other numbers. We can get proofs of their irrationality by using some of these connections, such as continued fraction expansions for both.
$\gamma$ may be thought of as a renormalized version of $\zeta(1)$, where $\zeta$ is the Riemann zeta function $\zeta(s) = \sum_{n=1}^\infty n^{-s}$.
$$\gamma = \lim_{s\to 1} \bigg(\zeta(s) - \frac{1}{s-1} \bigg)$$
At even integers, $\zeta(s)$ may be rewritten as a sum over nonzero integers, not just the positive integers. That's one explanation for why it is easier to get a handle on $\zeta(s)$ at even values (where it is a rational times $\pi^s$) than at positive odd integer values. See the answers to "Establishing zeta(3) as a definite integral and its computation."
There is some hope. Apéry proved that $\zeta(3)$ is irrational, and this can be related to proofs that other well known numbers are irrational. There are expressions for $\pi$, $\log 2$, $\zeta(3)$ as periods, definite integrals of algebraic functions on $[0,1]$. These can be used in a unified way to prove all of these are irrational (although it's still tricky for $\zeta(3)$), and there are conjectures about the possible rational or algebraic relations between periods. However, so far, $\gamma$ isn't known to be a period although it is an exponential period (as is $e$). No other values of $\zeta$ at positive odd integers are individually known to be irrational.