Luo, Rudnick, and Sarnak prove that if $\pi$ is a cuspidal automorphic representation of $\mathrm{GL}_n(\mathbb{A}_{\mathbb{Q}})$, then
$|\alpha_{j,\pi}(p)|\leq p^{\frac{1}{2}-\frac{1}{n^2+1}}$
for $$ |\alpha_{j,\pi}(p)|\leq p^{\frac{1}{2}-\frac{1}{n^2+1}} $$ for all $1\leq j\leq n$ and any unramified prime $p$. The ramification restriction was removed by Blomer and Brumley.
For $1\leq k\leq K$, let $\pi_k$ be a cuspidal automorphic representation of $\mathrm{GL}_{n_k}(\mathbb{A}_{\mathbb{Q}})$. Consider the isobaric sum $\Pi = \pi_1\boxplus \pi_2 \boxplus \cdots\boxplus \pi_K$. Then
$L(s,\Pi) = \prod_{k=1}^K L(s,\pi_k)$,
and $$ L(s,\Pi) = \prod_{k=1}^K L(s,\pi_k), $$ and a bound for the local roots of $\Pi$ at $p$ reduces to a combination of the bounds for each of the cuspidal constituents $\pi_k$. Here is a convenient and uniform (yet sub-optimal) bound: If $N$ is the maximum of the $n_k$'s, then
$|\alpha_{j,\Pi}(p)|\leq p^{\frac{1}{2}-\frac{1}{N^2+1}}$, $1\leq j\leq n_1+\cdots+n_K$.
If $$ |\alpha_{j,\Pi}(p)|\leq p^{\frac{1}{2}-\frac{1}{N^2+1}},\quad 1\leq j\leq n_1+\cdots+n_K. $$ If $\pi$ (resp. $\pi'$) is a cuspidal automorphic representation of $\mathrm{GL}_m(\mathbb{A}_{\mathbb{Q}})$ (resp. $\mathrm{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$), then one has
$|\alpha_{j,j',\pi\times\pi'}(p)|\leq p^{\frac{1}{2}-\frac{1}{m^2+1}+\frac{1}{2}-\frac{1}{n^2+1}}=p^{1-\frac{1}{m^2+1}-\frac{1}{n^2+1}}$ $$ |\alpha_{j,j',\pi\times\pi'}(p)|\leq p^{\frac{1}{2}-\frac{1}{m^2+1}+\frac{1}{2}-\frac{1}{n^2+1}}=p^{1-\frac{1}{m^2+1}-\frac{1}{n^2+1}} $$ where ($1\leq j\leq m$ and $1\leq j'\leq n$).
This follows from the explicit description of the local roots (including at the ramified primes) given by Brumley in the appendix to this paper.
ADDED: In the Selberg class, if you strictly adhere to the axioms, there is no notion of $\alpha_{j,\pi}(p)$. One has a Dirichlet series for $L(s)$ and for $\log L(s)$, but you are not guaranteed that the Dirichlet series for $L(s)$ factors as a product of local roots like
$\displaystyle \prod_{p}\prod_{j=1}^m(1-\alpha_{j,\pi}(p)p^{-s})^{-1}$. $$ \displaystyle \prod_{p}\prod_{j=1}^m(1-\alpha_{j,\pi}(p)p^{-s})^{-1}. $$