Here is a *heuristic* answer inspired by [this comment][1] of [Lucia][2].

First, let assume that the probabilty for an integer $n$ to be odd is $\frac{1}{2}$, and that the probabilty for $f(n)$ to be odd when $n$ is even (resp. odd) is also $\frac{1}{2}$. We will observe that (surprisingly) it is no more $\frac{1}{2}$ for $f^{\circ r}(n)$ when $r \ge 2$ (in some sense, the probability does not commute with the composition of $f$ with itself).  

- if $n$ and $m=f(n)$ are even: note that $\frac{n}{\sqrt{2}} = m+\theta$ (with $0 < \theta < 1$) so that $m=\frac{n}{\sqrt{2}}- \theta$, then $$f^{\circ 2}(n) = f(m) = \left \lfloor{\frac{m}{\sqrt{2}}} \right \rfloor = \left \lfloor{\frac{\frac{n}{\sqrt{2}}- \theta}{\sqrt{2}}} \right \rfloor = \left \lfloor \frac{n}{2} - \frac{\theta}{\sqrt{2}}\right \rfloor$$ but $\frac{n}{2}$ is even with probability $\frac{1}{2}$, so in this case, $f^{\circ 2}(n)$ is odd with probability $\frac{1}{2}$.  

- if $n$ is even and $m=f(n)$ is odd: $$f^{\circ 2}(n) = f(m) = \left \lfloor\sqrt{2}m \right \rfloor = \left \lfloor \sqrt{2}(\frac{n}{\sqrt{2}} - \theta) \right \rfloor = \left \lfloor n - \sqrt{2} \theta) \right \rfloor$$ but $n$ is even and the probability for $0<\sqrt{2} \theta<1$ is $\frac{\sqrt{2}}{2}$ (because $\theta$ is assumed statistically equidistributed on the open interval $(0,1)$), so $f^{\circ 2}(n)$ is odd with probability 
$\frac{\sqrt{2}}{2}$.    

- if $n$ is odd and $m=f(n)$ is even:  
$$f^{\circ 2}(n) = f(m) = \left \lfloor{\frac{m}{\sqrt{2}}} \right \rfloor = \left \lfloor{\frac{\sqrt{2}n-\theta}{\sqrt{2}}} \right \rfloor = \left \lfloor n - \frac{\theta}{\sqrt{2}} \right \rfloor  $$
but $n$ is odd and $0 < \frac{\theta}{\sqrt{2}}<1$, so $f^{\circ 2}(n)$ is even.  

- if $n$ is odd and $m=f(n)$ is odd:    
$$f^{\circ 2}(n) = f(m) = \left \lfloor \sqrt{2} m \right \rfloor = \left \lfloor \sqrt{2} (\sqrt{2}n-\theta) \right \rfloor = \left \lfloor 2n - \sqrt{2} \theta \right \rfloor $$
but $2n$ is even and the probability for $0<\sqrt{2} \theta<1$ is $\frac{\sqrt{2}}{2}$, so $f^{\circ 2}(n)$ is odd with probability $\frac{\sqrt{2}}{2}$.  

By combining these four cases together, we deduce that the probability for $f^{\circ 2}(n)$ to be odd is $$\frac{1}{2} \times \frac{1}{2} \times (\frac{1}{2} + \frac{\sqrt{2}}{2} + 0 + \frac{\sqrt{2}}{2}) = \frac{2\sqrt{2}+1}{8}$$  

By continuing in the same way, we get that the probability for $f^{\circ 3}(n)$ to be odd is:  
$$ \frac{1}{4} (\frac{1}{2}\frac{1}{2} + \frac{1}{2}\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2} + 1\frac{1}{2} + \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2})  = \frac{\sqrt{2}+7}{16}$$   

For $2 \le r \le 24$, we computed the probability $p_r$ for $f^{\circ r}(n)$ to be odd (see Appendix). It seems (experimentally) that $p_r$ converges to a number $\simeq 0.532288725 \simeq \frac{8+3\sqrt{2}}{23}$ by [Inverse Symbolic Calculator][3]. This leads to the following question/conjecture:    
$$\lim_{r \to \infty}p_r = \frac{8+3\sqrt{2}}{23} \ \  ?$$  

If so, consider the number $\alpha$ mentioned in the main post, then $$\alpha = 1-\frac{8+3\sqrt{2}}{23} = \frac{15-3\sqrt{2}}{23} \simeq 0.467711,$$ which matches with the computation in the main post. And next, we would have:  
$$ \delta = \frac{\sqrt{2}}{2^{\alpha}}= 2^{\frac{1}{2}-\alpha} = 2^{\frac{6\sqrt{2}-7}{46}} \simeq 1.022633$$

_________________
**Appendix**

*Computation*

    sage: for i in range(3,26):
    ....:     print(sq2(i))
    ....:
    [1/4*sqrt(2) + 1/8, 0.478553390593274]
    [1/16*sqrt(2) + 7/16, 0.525888347648318]
    [3/32*sqrt(2) + 13/32, 0.538832521472478]
    [15/64*sqrt(2) + 13/64, 0.534581303681194]
    [5/128*sqrt(2) + 61/128, 0.531805217280199]
    [39/256*sqrt(2) + 81/256, 0.531852847392776]
    [93/512*sqrt(2) + 141/512, 0.532269260352925]
    [51/1024*sqrt(2) + 473/1024, 0.532348527032254]
    [377/2048*sqrt(2) + 557/2048, 0.532303961432938]
    [551/4096*sqrt(2) + 1401/4096, 0.532283123258685]
    [653/8192*sqrt(2) + 3437/8192, 0.532285334012406]
    [3083/16384*sqrt(2) + 4361/16384, 0.532288843554459]
    [3409/32768*sqrt(2) + 12621/32768, 0.532289246647030]
    [7407/65536*sqrt(2) + 24409/65536, 0.532288816169701]
    [22805/131072*sqrt(2) + 37517/131072, 0.532288667983386]
    [24307/262144*sqrt(2) + 105161/262144, 0.532288700334941]
    [72761/524288*sqrt(2) + 176173/524288, 0.532288728736551]
    [159959/1048576*sqrt(2) + 331929/1048576, 0.532288729880941]
    [202621/2097152*sqrt(2) + 829741/2097152, 0.532288725958633]
    [639131/4194304*sqrt(2) + 1328713/4194304, 0.532288724978704]
    [1114081/8388608*sqrt(2) + 2889613/8388608, 0.532288725350163]
    [1825983/16777216*sqrt(2) + 6347993/16777216, 0.532288725570602]
    [5183461/33554432*sqrt(2) + 10530125/33554432, 0.532288725561857]


*Code*

    def sq2(n):
    	c=0
    	for i in range(2^n):
    		l=list(Integer(i).digits(base=2,padto=n))
    		if l[-1]==1:
    			cc=1/4
    			for j in range(n-2):
    				ll=[l[j],l[j+1],l[j+2]]
    				if ll==[0,0,0]:
    					cc*=1/2
    				if ll==[0,0,1]:
    					cc*=1/2
    				if ll==[0,1,0]:
    					cc*=(1-sqrt(2)/2)
    				if ll==[0,1,1]:
    					cc*=sqrt(2)/2
    				if ll==[1,0,0]:
    					cc*=1
    				if ll==[1,0,1]:
    					cc=0
    					break
    				if ll==[1,1,0]:
    					cc*=(1-sqrt(2)/2)
    				if ll==[1,1,1]:
    					cc*=sqrt(2)/2
    			c+=cc
    	return [c.expand(),c.n()]


  [1]: https://mathoverflow.net/questions/353524/the-irrational-numbers-%CE%B1-such-that-n-odd-and-m-n%CE%B1-odd-implies-m%CE%B1-odd#comment886622_353524
  [2]: https://mathoverflow.net/users/38624/lucia
  [3]: http://wayback.cecm.sfu.ca/cgi-bin/isc/lookup?number=0.532288725&lookup_type=simple