Of course the general answer to the question in the title is: not very simple.
I could not think of a better title, so let me explain my question in more detail.


I have a number field $E/\mathbb{Q}$, and a simple group $H/E$.
Let $G$ denote the Weil restriction of scalars of $H$ from $E$ to $\mathbb{Q}$.
Then $G$ is a $\mathbb{Q}$-simple group. 
The group $G\_{\ell} := G\_{\mathbb{Q}\_{\ell}}$ is not $\mathbb{Q}\_{\ell}$-simple in general.


*Short aside:* However, if $n = [E : \mathbb{Q}] \le 3$ there is always an $\ell$ such that $G\_{\ell}$ is $\mathbb{Q}\_{\ell}$-simple. (For a short proof: consider the Galois action of $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ on the $n$ factors of $G\_{\bar{\mathbb{Q}}}$. This action is transitive. Since $n \le 3$, there must therefore be an $n$-cycle in the action, and by Chebotarev there will be (infinitely many) primes $\ell$, such that $\mathrm{Gal}(\bar{\mathbb{Q}\_{\ell}}/\mathbb{Q}\_{\ell})$ acts transitively on the factors of $G\_{\bar{\mathbb{Q}\_{\ell}}}$. [Proof copied from D. Lombardo, arXiv: 1402.1478])


My question is about what happens if $n > 3$. (In my particular case $n \le 10$.) For example, I would be very happy, if there is always a prime $\ell$, so that all the $\mathbb{Q}\_{\ell}$-simple factors of $G\_{\ell}$ are not absolutely simple. Is something like that true?



How simple does a $\mathbb{Q}$-simple group remain after base change to $\mathbb{Q}_{\ell}$?
I assume that the question is: Is it true that there is always a prime $\ell$ such that $G\_\ell$ has no absolutely simple direct factors? 
The answer is YES. There are infinitely many such $\ell$.



> 
> **Claim.** Let $\Gamma$ be a finite group acting transitively on a finite set $S$ of cardinality $n>1$. 
>  Then $\Gamma$ has a cyclic subgroup $\Delta\subset \Gamma$ acting on $S$ without fixed points.
> 
> 
> 


**Proof of the claim** (due to Guntram). Assume the contrary. 
Then any element $\gamma$ of $\Gamma$ is contained in the stabilizer $\Gamma\_s$ of some point $s\in S$,
hence $\Gamma=\bigcup\_s\Gamma\_s$.
Since $\Gamma$ acts transitively on $S$, each subgroup $\Gamma\_s$ is of index $n$ in $\Gamma$. 
Since $n>1$ and $1\in \Gamma\_s$ for all $s\in S$, we obtain that
$$ \#\bigcup\_s\Gamma\_s\le 1+ n\cdot (\#\Gamma\_s-1)<\#\Gamma, $$
contradiction.


Now let $S$ denote the set of embeddings of $E$ into $\overline{\mathbb{Q}}$. The Galois group
of $\overline{\mathbb{Q}}$ over ${\mathbb{Q}}$ acts on $S$ via a finite quotient group $\Gamma$.
Let $\Delta=\langle\gamma\rangle$ be as in the Claim. Then
by Chebotarev's density theorem there exist infinitely many primes $\ell$ unramified in $E$ 
such that the corresponding Frobenius is conjugate to $\gamma$.
For such an $\ell$, the tensor product $E\otimes\_{\mathbb{Q}} {\mathbb{Q}}\_\ell$ 
does not contain ${\mathbb{Q}}\_\ell$ as a direct summand, and therefore, 
$G\_\ell$ has no absolutely simple direct factors.


