Indeed the radius of convergence of the function $f$ is zero, which means that the series
$$ \sum_{k=0}^{\infty} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k $$
diverges for all $z \neq 0$. Although this can be directly verified by using Stirling's approximation, a more fundamental reason for this is that $0$ is a branch point of $f(z)$. Consequently, there is no way $f(z)$ extends to an analytic function about $0$. For instance, the closed-form
$$ f(z) = \sqrt{\frac{3}{2\pi z}} \, e^{3/4z} K_{1/4}(3/4z) \tag{*} $$
in Claude Leibovici's answer has the branch cut $(-\infty, 0]$, as seen from the domain coloring plot of $\text{(*)}$ on the rectangle with the corners $\pm 5 \pm 5i$:
That said, your best hope for an asymptotic expansion of polynomial form is
$$ f(z) = \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k + \mathcal{O}(z^N) \qquad \text{as } z \to 0 \tag{1} $$
within Stolz angle for each fixed $N \geq 0$.
Addendum. Here is a proof that $\text{(1)}$ holds at least within the region $\operatorname{Re}(z) \geq 0$, although I suspect that this holds within larger sectors.
Note that for each fixed $N$, Taylor's theorem tells that for $\operatorname{Re}(z) \geq 0$,
\begin{align*}
&f(z) - \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k \\
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl[ \exp\left(-\frac{zy^4}{24}\right) - \sum_{k=0}^{N-1} \frac{1}{k!} \left( -\frac{zy^4}{24} \right)^k \Biggr] e^{-\frac{y^2}{2}} \, \mathrm{d}y \\
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl( \int_{0}^{1} \frac{z^N (1-t)^{N-1}}{(N-1)!} \left(-\frac{y^4}{24}\right)^N \exp\left(-\frac{zty^4}{24}\right) \, \mathrm{d}t \Biggr) e^{-\frac{y^2}{2}} \, \mathrm{d}y,
\end{align*}
where the last line follows from the Lagrange form of the remainder. Taking absolute value,
\begin{align*}
&\left| f(z) - \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k \right| \\
&\leq \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl( \int_{0}^{1} \frac{\left| z \right|^N (1-t)^{N-1}}{(N-1)!} \frac{y^{4N}}{24^N} \, \mathrm{d}t \Biggr) e^{-\frac{y^2}{2}} \, \mathrm{d}y \\
&= \frac{(4N-1)!!}{N!24^N} \left|z\right|^N.
\end{align*}
This proves that the error term in $\text{(*)}$ is indeed $\mathcal{O}(z^N)$, at least within the region $\operatorname{Re}(z) \geq 0$.
Best Answer
To get an expansion for $x>>1$, let's first make substitution $y = x^{-\frac14}t$ in the integral: \begin{align} f(x) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{t^4}{24} - \frac{t^2}{2\sqrt{x}}} \frac{dt}{\sqrt[4]{x}} = \frac{x^{-\frac14}}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{t^4}{24}} \sum_{n=0}^\infty \frac{1}{n!}\left(- \frac{t^2}{2\sqrt{x}}\right)^n dt \\ &= \sum_{n=0}^\infty \frac{(-1)^nx^{-\frac14-\frac{n}{2}}}{\sqrt{2\pi} 2^n n!} \int_{-\infty}^\infty t^{2n}e^{-\frac{t^4}{24}} dt \end{align} For these integrals we have, with substitution $t = \sqrt[4]{24u} $ \begin{align} \int_{-\infty}^\infty t^{2n}e^{-\frac{t^4}{24}} dt &= 2 \int_0^\infty t^{2n}e^{-\frac{t^4}{24}} dt = \\ &= 2 (24)^{\frac{2n+1}{4}}\int_0^\infty u^{\frac{2n-3}{4}}e^{-u} du = \\ &= 2 (24)^{\frac{2n+1}{4}} \Gamma(\frac{2n+1}{4}) \end{align} where $\Gamma$ is the Euler's gamma function. In total, we get $$ f(x) = \sum_{n=0}^\infty \frac{(-1)^n \Gamma(\frac{n}{2}+\frac14)}{\sqrt{2\pi} 2^{n-1} n!} (\frac{x}{24})^{-\frac{2n+1}{4}} $$ It can be shown that this series is convergent for all $x>0$.