One way to see it is to look at the formula for the Fourier transform $\hat{f_\lambda}(\xi)$:
$$
\hat{f_\lambda}(\xi) = \int f_\lambda(x) e^{-2\pi ix\xi}\,dx = \int e^{-\pi x^2} e^{i\pi(\lambda x^2 -2x\xi)}\,dx.
$$
The phase function $\phi_\xi(x) = \lambda x^2-2x\xi$ has a critical point at $x = \frac{\xi}{\lambda}$. Expanding $\phi_\xi(\frac{\xi}{\lambda}+h) = \phi_\xi(\frac{\xi}{\lambda}) + \frac{h^2}{2}\phi_\xi''(\frac{\xi}{\lambda}) = -\frac{\xi^2}{\lambda}+\lambda h^2$, we have
$$
\hat{f_\lambda}(\xi) = e^{-i\pi \frac{\xi^2}{\lambda}}\int e^{-\pi (\frac{\xi}{\lambda}+h)^2} e^{i\pi\lambda h^2}\,dh.
$$
If $|h|\gg \frac{1}{\sqrt \lambda}$, then $e^{i\pi\lambda h^2}$ will be oscillating very fast, so the contribution to the integral appearing above from $|h|\gg \frac{1}{\sqrt \lambda}$ will be very small. When $|\xi|\gg\lambda$ and $|h|\lesssim \frac{1}{\sqrt{\lambda}}$, the amplitude $e^{-\pi(\frac{\xi}{\lambda}+h)^2}$ will be very small. While when $|\xi|\lesssim \lambda$ and $|h|\lesssim \frac{1}{\sqrt{\lambda}}$, the integrand $e^{-\pi (\frac{\xi}{\lambda}+h)^2} e^{i\pi\lambda h^2}$ will be approximately $1$.
Putting these observations together, the integral $\int|\hat{f_\lambda}(\xi)|^{q}\,d\xi$ is approximately
\begin{align*}
\int|\hat{f_\lambda}(\xi)|^{q}\,d\xi\approx \int_{|\xi|\lesssim \lambda}|\int_{|h|\lesssim\frac{1}{\sqrt\lambda}}1\,dh|^{q}\,d\xi \lesssim \lambda^{1-\frac{q}{2}}.
\end{align*}
Taking the $q$th root gives $\|\hat{f_\lambda}\|_{q} \le c \lambda^{\frac{1}{q}-\frac{1}{2}}$.
