This result (and much more) is in Exposé IX *Modèles de Néron et monodromie* by A.Grothendieck (in SGA7), and more precisely in section 11.1. In particular what you want is exactly Proposition 11.2 thereof.

The proof is not hard but it also far from being obvious (to me, anyway). Let me recall it briefly. A separate, quasi-finite scheme $X$ over $\mathbb Z_{p}$ decomposes as a disjoint union of a finite part $X^{f}$ which has same special fiber as $X$ with a scheme $X'$ with zero special fiber. Denote by $\mathcal E$ the Néron model of $E$. The composition of $\mathcal E(\bar{\mathbb Q}_{p})^{f}\rightarrow (\mathcal E\times_{\mathbb Z_{p}}\mathbb F_{p})(\bar{\mathbb F}_{p})\rightarrow \Phi$ is surjective and its kernel is $\mathcal E(\bar{\mathbb Q}_{p})^{f,\circ}$. As multiplication by $\ell^n$ is surjective on $\mathcal E^{\circ}(\bar{\mathbb Q}_{p})^{f}$, the Snake Lemma gives an isomorphism $\Phi[\ell^{n}]\simeq\mathcal E[\ell^n]^{f}/\mathcal E[\ell^n]^{f,\circ}$. The image of $(T_\ell E)^f\rightarrow E[\ell^n]^f$ is $\ell$-divisible so in $E[\ell^n]^{f,\circ}$ as $\Phi$ is finite and coincides with this subspace by surjectivity of multiplication by $\ell^n$. Hence $(T_\ell E)^{f}\otimes\mathbb Z/\ell^{n}\mathbb Z\simeq \mathcal E[\ell^n]^{f,\circ}$.

Finally, the finite part of $\mathcal E[\ell^n]$ is given by the invariants under the inertia group $I_p$. Putting all this together, we get the following explicit description of the $\ell^n$-torsion of the component group and the result
$$\Phi[\ell^n]\simeq\frac{(T_{\ell}E\otimes\mathbb Z/\ell^{n}\mathbb Z)^{I_p}}{(T_{\ell}E)^{I_p}\otimes\mathbb Z/\ell^{n}\mathbb Z}.$$