Since $\lambda$ is holomorphic on the unit $q^{1/2}$-disc with singularities on the boundary, the power-series development has radius of convergence equal to one.

Sorry about the one-sentence answer. Here is an expanded version that uses the following facts: The modular function $\lambda$ is holomorphic on the upper half-plane, invariant under the action of $\Gamma(2)$, and non-singular at infinity.

From the invariance under the subgroup of $\Gamma(2)$ generated by $\binom{12}{01} = (z \mapsto z + 2)$, we can conclude that $\lambda$ has a $q$-expansion in powers of $q^{1/2}$, and from regularity at infinity, these powers are nonnegative. In fact, the periodicity combined with holomorphicity implies the map $\lambda: \mathcal{H} \to \mathbb{C}$ factors as a composite of $(\tau \mapsto e^{i \pi \tau}): \mathcal{H} \to D_{q^{1/2}}$ followed by $\bar{\lambda}: D_{q^{1/2}} \to \mathbb{C}$, where $D_{q^{1/2}}$ is the open unit disc of radius one with coordinate $q^{1/2}$, and $\bar{\lambda}$ is a holomorphic function uniquely defined by this factorization. It's power series expansion at $q^{1/2}=0$ is given by the $q$-expansion of $\lambda$, and because the function is holomorphic on the open unit disc, the radius of convergence of the series is at least one.

To show that the radius of convergence is exactly one, we note that $\lambda$ is singular on a $\Gamma(2)$-orbit of cusps, and these cusps are taken to a dense subset of the unit circle under the $\tau \mapsto e^{i \pi \tau}$ map. The fact that $\lambda$ is singular at a nonempty set of cusps follows from the fact that $\lambda$ factors through the $\Gamma(2)$-quotient map $\mathcal{H} \twoheadrightarrow Y(2)$, the compactification $Y(2) \hookrightarrow X(2)$, and an analytic isomorphism $X(2)\overset{\simeq}{\longrightarrow} \mathbb{P}^1$. Since $Y(2)$ is dense in $X(2)$, the composite has dense image but the fact that $\lambda$ is holomorphic and nonconstant on $\mathcal{H}$ implies the image approaches but does not include infinity.

Regarding more precise asymptotic estimates for coefficients, if you want to apply the Rademacher circle method to get asymptotics of a form of level greater than one (such as $\lambda$), you need to work with the singularities at all cusps. There is an extremely brief description of this method in section 5 of Borcherds's paper Automorphic forms on $O_{s+2,2}(\mathbf{R})$ and infinite products. Alternatively, you may turn $\lambda$ into a vector-valued form of level one with respect to a representation of $\Gamma(1)$.