We show the leading term of the asymptotic expansion of the sum is
\begin{align*}
\sum_{1\leq k \leq n/3 } \binom{2k}{k}\binom{n-2k-1}{k-1} \color{blue}{\sim \frac{2^{n-1}}{\sqrt{n\pi}}}\tag{1}
\end{align*}
Denoting the sum with $a_n$ we get
\begin{align*}
(a_n)_{n\geq 0}=\{1,0,0,2,2,2,8,14,20,46,92,158,\ldots\}
\end{align*}
which is OEIS A360309. We can find there a generating function
\begin{align*}
{\color{blue}{A(z)}}=\sum_{n\geq 0}a_nz^n=\frac{1}{\sqrt{1-\frac{4z^3}{1-z}}}
\color{blue}{=\frac{\sqrt{1-z}}{\sqrt{1-z-4z^3}}}\tag{2}
\end{align*}
A singularity analysis of (2) reveals the leading term of the asymptotic expansion (and more terms if we like). At first we locate the dominant singularity of (2) which is the one nearest to the origin. Since the zeros of the denominator
\begin{align*}
\sqrt{1-z-4z^3}=\sqrt{(-4)(z-z_0)(z-z_1)(z-z_2)}\tag{3}
\end{align*}
are the branch points
\begin{align*}
z_0=\frac{1}{2},\quad z_1=-\frac{1}{4}\left(1+i\sqrt{7}\right),\quad z_2=-\frac{1}{4}\left(1-i\sqrt{7}\right)
\end{align*}
with
\begin{align*}
\left|z_0\right|=\frac{1}{2}\qquad\text{and}\qquad\left|z_{1}\right|=\left|z_{2}\right|=\frac{\sqrt{2}}{2}
\end{align*}
the dominant singularity determining the asymptotic behaviour of $(a_n)_{n\geq 0}$ is $z_0=\frac{1}{2}$. Expanding all terms of the denominator as well as the numerator at $z=1/2$ gives with some help of Wolfram Alpha
\begin{align*}
\frac{1}{\sqrt{z-z_0}}&=\color{blue}{\frac{1}{\sqrt{z-\frac{1}{2}}}}\tag{4}\\
\frac{1}{\sqrt{z-z_1}}&=\frac{1}{\sqrt{1+\frac{1}{4}\left(1+i\sqrt{7}\right)}}
={\color{blue}{\frac{2}{\sqrt{3+i\sqrt{7}}}}}+\mathcal{O}\left(z-\frac{1}{2}\right)\tag{5.1}\\
\frac{1}{\sqrt{z-z_2}}&=\frac{1}{\sqrt{1+\frac{1}{4}\left(1-i\sqrt{7}\right)}}
={\color{blue}{\frac{2}{\sqrt{3-i\sqrt{7}}}}}+\mathcal{O}\left(z-\frac{1}{2}\right)\tag{5.2}\\
\sqrt{1-z}&={\color{blue}{\frac{1}{\sqrt{2}}}}+\mathcal{O}\left(z-\frac{1}{2}\right)\tag{5.3}\\
\end{align*}
The leading term of the asymptotic expansion is determined by the term $\frac{1}{\sqrt{z-\frac{1}{2}}}$ multiplied by the constant $\frac{1}{\sqrt{-4}}$ from (3) and the blue marked constants from (5.1) to (5.3). We obtain
\begin{align*}
\frac{1}{\sqrt{(-4)\left(z-\frac{1}{2}\right)}}
\cdot\frac{2}{\sqrt{3+i\sqrt{7}}}
\cdot\frac{2}{\sqrt{3-i\sqrt{7}}}
\cdot\frac{1}{\sqrt{2}}=\color{blue}{\frac{1}{2\sqrt{1-2z}}}\tag{6}
\end{align*}
Asymptotic expansion:
We follow chapter VI in Analytic Combinatorics by P. Flajolet and R. Sedgewick and get from Theorem VI.1 an asymptotic expansion of
\begin{align*}
[z^n]f(z)=(1-z)^{-\alpha}\qquad\qquad\alpha\in\mathbb{C}\setminus\mathbb{Z}_{\leq 0}\tag{7}
\end{align*}
We take terms up to order $1$ and get
\begin{align*}
[z^n]f(z)\sim\frac{n^{\alpha-1}}{\Gamma(\alpha)}\left(1+\mathcal{O}\left(\frac{1}{n}\right)\right)\tag{8}
\end{align*}
We need from (8) the expansion
\begin{align*}
\color{blue}{[z^n](1-z)^{-\frac{1}{2}}}&\sim\frac{n^{-\frac{1}{2}}}{\Gamma\left(\frac{1}{2}\right)}
\left(1+\mathcal{O}\left(\frac{1}{n }\right)\right)\\
&\color{blue}{=\frac{1}{\sqrt{n\pi}}\left(1+\mathcal{O}\left(\frac{1}{n }\right)\right)}\tag{9}\\
\end{align*}
Note, that (7) is given in standard form, but we have something like $(1-qz)^{-\alpha}$ instead. We overcome this by recalling a Taylor series expansion and noting that
\begin{align*}
[z^n]f(z)&=q^{-n}[z^n]f\left(qz\right)\\
[z^n]f(qz)&=q^n[z^n]f(z)\\
[z^n]\frac{1}{\sqrt{1-2z}}&=2^n[z^n]\frac{1}{\sqrt{1-z}}\tag{10}
\end{align*}
We finally get from (6), (9) and (10)
\begin{align*}
\color{blue}{\frac{1}{2\sqrt{1-2z}}\sim
\frac{2^{n-1}}{\sqrt{n\pi}}\left(1+\mathcal{O}\left(\frac{1}{n }\right)\right)}
\end{align*}
as stated in (1) and in accordance with the answer from @ClaudeLeibovici.
Add-on: For the sake of completeness we also provide a derivation of the generating function $A(z)$. We use the coefficient of operator $[z^n]$ to denote the coefficient of a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\tag{11}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{1\leq k \leq n/3 }}&{\color{blue}{ \binom{2k}{k}\binom{n-2k-1}{k-1}}}\\
&=\sum_{k=0}^{\infty}\binom{2k}{k}\binom{n-2k-1}{n-3k}\tag{12.1}\\
&=\sum_{k=0}^{\infty}\binom{2k}{k}\binom{-k}{n-3k}(-1)^{n-3k}\tag{12.2}\\
&=\sum_{k=0}^{\infty}\binom{2k}{k}[z^{n-3k}]\frac{1}{(1-z)^k}\tag{12.3}\\
&=[z^n]\sum_{k=0}^{\infty}\binom{2k}{k}\left(\frac{z^3}{1-z}\right)^k\tag{12.4}\\
&=[z^n]\sum_{k=0}^{\infty}\binom{-1/2}{k}\left(-\frac{4z^3}{1-z}\right)^k\tag{12.5}\\
&\,\,\color{blue}{=[z^n]\frac{1}{\sqrt{1-\frac{4z^3}{1-z}}}}
\end{align*}
in accordance with (2).
Comment:
In (12.1) we use the binomial identitiy $\binom{p}{q}=\binom{p}{p-q}$ and we also extend the range of summation without any change as we only add zeros.
In (12.2) we use $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
In (12.3) we use the coefficient of operator as in (11) and a binomial series expansion.
In (12.4) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
In (12.5) we use the binomial identity $\binom{2k}{k}=(-1)^k4^k\binom{-1/2}{k}$ of the central binomial coefficient and perform again a binomial series expansion in the last line.
