If $\zeta(s)=0$ with $\Re(s)=\frac{1}{2}$, is then $|\hat{\zeta}(s,3)|^2=\frac{1}{2}$?
This is just a simple calculation. Note that $$ \sum_{n=k}^{\infty} \frac{1}{2^{n+1}} \binom{n}{k} =1 $$ for any non-negative integer $k$, which immediately gives $$ \frac{1}{1-2^{1-s}} \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^s} \binom{n}{k} = \frac{1}{1-2^{1-s}}\sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^s} = \zeta(s). $$
Similarly, $$ {\hat \zeta}(s,3) = \frac{1}{1-2^{1-s}} \sum_{k=0}^{\infty} (-1)^k K(k,3)^s = \frac{1}{1-2^{1-s}} \sum_{k=0}^{\infty} \frac{(-1)^k (k,3)^s}{(k+3)^s}. $$ Separating out terms with $3|k$, we may rewrite the sum over $k$ above as \begin{align*} &\sum_{k=0}^{\infty} \frac{(-1)^k}{(k+3)^s} + (3^s-1)\sum_{k=0}^{\infty} \frac{(-1)^{3k}}{(3k+3)^s}\\ = &-1 + \frac{1}{2^s} +\sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^s} + \frac{3^s-1}{3^s} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^s}. \end{align*} Therefore $$ {\hat \zeta}(s,3) = \frac{-1+2^{-s}}{1-2^{1-s}} + \zeta(s) + \frac{3^s-1}{3^s} \zeta(s). $$ Now evaluate this at a zero of $\zeta(s)$ on the critical line (or anywhere else).
The intent of this post is to share some results as requested in the comments above and assumes the following definitions from the original question above.
(1) $\quad\zeta(s)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{1-2^{1-s}}\sum_\limits{n=0}^N\frac{1}{2^{n+1}}\sum_\limits{k=0}^n(-1)^k\left( \begin{array}{c} n \\ k \\ \end{array} \right)\frac{1}{(k+1)^s}\right)$
(2) $\quad\hat{\zeta}(s,a)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{1-2^{1-s}}\sum_\limits{n=0}^N\frac{1}{2^{n+1}}\sum_\limits{k=0}^n(-1)^k\left( \begin{array}{c} n \\ k \\ \end{array} \right)K(k,a)^{s}\right),\qquad K(a,b)=\frac{\gcd(a,b)}{a+b}$
The following figure illustrates formula (2) above for $|\hat{\zeta}(s,3)|$ plotted along the critical line $s=\frac{1}{2}+i\ t$. The red discrete portion of the plot represents the evaluation of formula (2) for $|\hat{\zeta}(s,3)|$ at the first 10 non-trivial zeta zeros in the upper half plane. In the plot in the figure below formula (2) was evaluated using an upper limit of $N=400$.
Figure (1): Illustration of formula (2) for $|\hat{\zeta}(s,3)|$ evaluated along the critical line
The following conjectured formulas based on my Questions related to globally convergent formulas for the Dirichlet eta function $\eta(s)$ seem to evaluate considerably faster than formulas (1) and (2) above and perhaps provide some additional insight.
(3) $\quad\zeta(s)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{\left(1-2^{1-s}\right)\ 2^{N+1}}\sum\limits_{n=0}^N (-1)^n\ \frac{1}{(n+1)^s}\sum\limits_{k=0}^{N-n} \left( \begin{array}{c} N+1 \\ N-n-k \\ \end{array} \right)\right)$
(4) $\quad\hat{\zeta}(s,a)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{\left(1-2^{1-s}\right)\ 2^{N+1}}\sum\limits_{n=0}^N (-1)^n\ K(n,a)^{s}\sum\limits_{k=0}^{N-n} \left( \begin{array}{c} N+1 \\ N-n-k \\ \end{array} \right)\right)$
The following figure illustrates formula (4) above for $|\hat{\zeta}(s,3)|$ plotted along the critical line $s=\frac{1}{2}+i\ t$. The red discrete portion of the plot represents the evaluation of formula (4) for $|\hat{\zeta}(s,3)|$ at the first 10 non-trivial zeta zeros in the upper half plane. In the plot in the figure below, formula (4) was evaluated using an upper limit of $N=400$.
Figure (2): Illustration of formula (4) for $|\hat{\zeta}(s,3)|$ evaluated along the critical line
Formula (5) below is from the accepted answer to this question posted by Lucia.
(5) $\quad\hat{\zeta}(s,3)=\frac{2^{-s}-1}{1-2^{1-s}}+\zeta(s)+\frac{3^s-1}{3^s} \zeta(s)$
The following figure illustrates formula (5) above for $|\hat{\zeta}(s,3)|$ and $\left|\frac{2^{-s}-1}{1-2^{1-s}}\right|$ in blue and orange respectively where both functions are evaluated along the critical line $s=\frac{1}{2}+i\ t$. The red discrete portion of the plot represents the evaluation of formula (5) for $|\hat{\zeta}(s,3)|$ at the first 10 non-trivial zeta zeros in the upper half plane.
Figure (3): Illustration of formula (5) for $|\hat{\zeta}(s,3)|$ evaluated along the critical line