Let $\omega$ be a primitive $d$ root of unity. Then define
$$\rm a:=\frac{1}{d}\sum_{j=0}^{d-1} \omega^{\,jk}=\begin{cases}1 & \rm k\equiv 0 ~\bmod d \\ 0 & \rm otherwise \end{cases}=\delta_k$$
To see the equality, first check the $\rm k\equiv0$ case, then the $\rm k\equiv 1$ case by symmetry ($\rm a=\omega \cdot a$), then notice the map $\rm j\mapsto \omega^{jk}$ sends the residues modulo $\rm d$ to the $\rm d/gcd(d,k)$th roots of unity, and it is specifically a $\rm \gcd(d,k)$-to-$1$ map, for any nonzero $\rm k$ (hence $\rm a=0$, going mod $\rm d/(d,k)$). (Also see the Kronecker delta function for a definition of the RHS.)
Using the binomial theorem again with interchange of summation, we have
$$\rm \begin{array}{c c}\frac{1}{d}\sum_{j=0}^{d-1} \big(1+\omega^j\big)^{nd} & \rm =\frac{1}{d}\sum_{j=0}^{d-1}\sum_{k=0}^{nd}\binom{nd}{k} \omega^{\,jk} \\
& \rm =\sum_{k=0}^{nd}\binom{nd}{k}\frac{1}{d}\sum_{j=0}^{d-1}\omega^{\,jk} \\
& \rm =\sum_{k=0}^{nd}\binom{nd}{k}\delta_k \\
& \rm =\sum_{\ell=0}^n \binom{nd}{n\ell}. \end{array}$$
Observe that the first leading term will be with $\omega=1$. Set $\rm \omega=e^{2\pi i/d}$; ordering $|1+\omega^j|$ by magnitude, we find the $(v+1)$th leading term is (for $v\le d/2$)
$$\begin{array}{c c} \rm \frac{(1+\omega^v)^{nd}+(1+\omega^{-v})^{nd}}{d} & = \rm
\frac{(\omega^{v/2}+\omega^{-v/2})^{nd}(\omega^{vnd/2}+\omega^{-vnd/2})}{d} \\
& =\rm \frac{2}{d}\big(2\cos(2\pi v/d) \big)^{nd}\cos(\pi vnd)\end{array}$$
Similar considerations lead to an offset modular generalization:
$$\rm \frac{1}{d}\sum_{j=0}^{d-1}\omega^{-j\,r} \big(1+\omega^j\big)^{nd} =\sum_{\ell=0}^n \binom{nd}{n\ell+r}.$$
Best Answer
Generally similar sums can be evaluated using the Beta function: $$ B(x+1,y+1)=\int_0^1 t^{x}(1-t)^{y}dt=\frac{\Gamma(x+1)\Gamma(y+1)}{\Gamma(x+y+2)}= \frac{x!y!}{(x+y+1)!}=\frac{1}{x+y+1}\binom{x+y}x^{-1}. $$
Applying this in your case ($x=n-k,y=n+k$) one has: $$ \binom{2n}{n-k}^{-1}=(2n+1)\int_0^1 t^{n-k}(1-t)^{n+k}dt $$ or $$\begin{align} \sum_{n=k}^\infty\binom{2n}{n-k}^{-1} &=\sum_{n=k}^\infty(2n+1)\int_0^1 t^{n-k}(1-t)^{n+k}dt\\ &=\int_0^1 dt \sum_{n=k}^\infty(2n+1)t^{n-k}(1-t)^{n+k}\\ &=\int_0^1 \frac{(1-t)^{2k}[2+(2k-1)(1-t+t^2)]}{(1-t+t^2)^2}dt\tag1 \end{align}$$
It can be shown that the integral $(1)$ is a sum of a rational number and a multiple of $\dfrac\pi{9\sqrt3}$.
Indeed: $$ \frac{(1-t)^{2k}[2+(2k-1)(1-t+t^2)]}{(1-t+t^2)^2}=Q_k(t)+\frac{A_k^0+A_k^1t+A_k^2t^2+A_k^3t^3}{(1-t+t^2)^2},\tag2 $$ where both coefficients of the polynomial $Q(t)$ and $A^0,A^1,A^2,A^3$ are integer numbers. The integral of $Q(t)$ is obviously a rational number and $I_r= \int_0^1\frac{t^r\,dt}{(1-t+t^2)^2}$ can be evaluated as: $$ I_0=\frac23+\frac49\dfrac\pi{\sqrt3};\quad I_1=\frac13+\frac29\dfrac\pi{\sqrt3};\quad I_2=-\frac13+\frac49\dfrac\pi{\sqrt3};\quad I_3=-\frac23+\frac59\dfrac\pi{\sqrt3}.\quad $$
Thus the irrational term can be written as: $$ C_k\frac\pi{9\sqrt3}\quad\text{with}\quad C_k=4A_k^0+2A_k^1+4A_k^2+5A_k^3. $$
Moreover the term can be evaluated explicitly using the following table: $$ \begin{array}{c|c|c|c|c|c} k\mod 3& A_k^0&A_k^1&A_k^2&A_k^3&C_k\\ \hline 0&+1-2x&+1+6x&-1-6x&+0+4x&2\\ 1&+2+4x&-5-6x&+3+6x&-1-2x&\hphantom{-1}5+18x\\ 2&-3-2x&2&0&-1-2x&-13-18x\\ \hline \end{array},\tag3 $$ with $x=\left\lfloor\dfrac k3\right\rfloor$, so that $C_k=2,5,-13,2,23,-31,2,41,-49,2,\dots$ for $k=0,1,2,3,4,5,6,7,8,9,\dots$.
The expression $(3)$ can be proved in the following way:
Let $$ P_k(t)=(1-t)^{2k}[2+(2k-1)(1-t+t^2)];\quad R_k(t)=A_k^0+A_k^1t+A_k^2t^2+A_k^3t^3.\\ $$
Then we have from (2): $$R_k(t_\pm)=P_k(t_\pm);\quad R'_k(t_\pm)=P'_k(t_\pm),\tag4$$ where $$ t_\pm=e^{\pm\frac{i\pi}3} $$ are the roots of the polynomial $t^2-t+1$.
Explicitly (4) amounts to the system of four linear equations: $$\begin{align} A_k^0+A_k^1t_\pm+A_k^2t_\pm^2+A_k^3t_\pm^3&=2t_\pm^{-2k}\\ A_k^1+2A_k^2t_\pm+3A_k^3t_\pm^2&=(1-2k-2t_\pm)t_\pm^{-2k}\\ \end{align}, $$ which solutions are given by (3).