Got numerical support that for odd $n$, $\zeta(n)$ might be
expressed in terms of the derivatives of $\zeta(\frac12)$.

Based on *More Zeta Functions for the Riemann Zeros*, Andre Voros, p.12, Table 3:

Conjecture:  For odd $n$, 
$$\zeta(n) = \left(\frac{2}{(n-1)!} (\log(|\zeta|)^{(n)} (\frac12) - 2^n \beta(n))\right)/(2^n-1)$$

$\beta(n)$ is Dirichlet beta function and it is a rational multiple of $\pi^n$
for odd $n$.
The derivative can be expressed in terms of $\zeta(\frac12),\zeta^{(k)}(\frac12)$

For $n=3$ get numerical support for:

$$\zeta(3) = (-\zeta'''(\frac12)/|\zeta(\frac12)| -3 \zeta''(\frac12) \zeta'(\frac12)/|\zeta(\frac12)|^2 -2 \zeta'(\frac12)^3/|\zeta(\frac12)|^3- \pi^3 / 4)/7 $$

The last equality holds with precision $10^4$ decimal digits.
One can eliminate the first derivative since there is closed form
for $\zeta'(\frac12)/\zeta(\frac12)$

>Is this result true?

sage/mpmath code in case of typos of the latex.

    #run in sage
    
    import mpmath
    from mpmath import mpf
    mpmath.mp.pretty=True
    
    def zeta3test():

        n=3
        mpmath.mp.dps=10^3
        zeta3=mpmath.zeta(3)
    
        Pi=mpmath.pi
        gamma=mpmath.euler
        z12=mpmath.zeta(1/mpmath.mpf(2))
        z1=mpmath.zeta(1/mpmath.mpf(2),derivative=1)
        z2=mpmath.zeta(1/mpmath.mpf(2),derivative=2)
        z3=mpmath.zeta(1/mpmath.mpf(2),derivative=3)
    
        # eliminate the first derivative
    
        #rh0=1/32*(72*Pi*mpmath.log(2)*mpmath.log(Pi)*z12+144*gamma*mpmath.log(2)*mpmath.log(Pi)*z12+72*mpmath.log(2)*z12*gamma*Pi+24*mpmath.log(Pi)*z12*gamma*Pi-144*z2*mpmath.log(2)-48*z2*mpmath.log(Pi)+216*mpmath.log(2)^3*z12+8*mpmath.log(Pi)^3*z12-48*z2*gamma-24*z2*Pi+8*z12*gamma^3+z12*Pi^3+72*mpmath.log(2)*z12*gamma^2+18*mpmath.log(2)*z12*Pi^2+24*mpmath.log(Pi)*z12*gamma^2+6*mpmath.log(Pi)*z12*Pi^2+216*mpmath.log(2)^2*mpmath.log(Pi)*z12+72*mpmath.log(2)*mpmath.log(Pi)^2*z12+216*gamma*mpmath.log(2)^2*z12+24*gamma*mpmath.log(Pi)^2*z12+108*Pi*mpmath.log(2)^2*z12+12*Pi*mpmath.log(Pi)^2*z12+32*z3+12*z12*gamma^2*Pi+6*z12*gamma*Pi^2)/z12
        z12a=mpmath.fabs(z12)
        rh1= -z3/z12a -3*z2*z1/z12a**2 -2* z1**3/z12a**3
    
        #print 'rh',mpmath.chop(rh0-rh1)
        #rhs= (mpmath.diff( lambda y:  mpmath.log(mpmath.fabs(mpmath.zeta(y))),1/2,n) - mpmath.pi^3 / 4 )/(7)
        rhs= (rh1 - mpmath.pi**3 / 4 )/(7)
        print mpmath.chop(zeta3-rhs)
    
    def conjecture1(n):
        """
        
        voros, p. 12
        """
        assert n%2==1
        a1= mpmath.zeta(n)
        a2= (2/factorial(n-1) * mpmath.diff( lambda y:  mpmath.log(mpmath.fabs(mpmath.zeta(y))),1/2,n) - 2**(n) * dirichletbeta(n))/(2**n-1)
        print mpmath.chop(a1-a2)
    
    
    def dirichletbeta(s):
        """
        dirichlet beta
        """
        return 4**(-s) * (mpmath.hurwitz(s,1/4)-mpmath.hurwitz(s,3/4))