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$\DeclareMathOperator\Res{Res}\DeclareMathOperator\Gal{Gal}$As noted, the answer is yes. Let $G$ be a simple algebraic group (meaning $F$-simple) over $F$.

Assume first that $F$ is Galois over $\mathbb{Q}$. Then $(\Res_{F/\mathbb{Q}}G)_{F}\simeq G_{1}\times\cdots\times G_{r}$, where the $G_{i}$ are the conjugates of $G$ under $\Gal(F/\mathbb{Q})$. If $H$ is a smooth connected normal algebraic subgroup of $\Res_{F/\mathbb{Q}}G$, then $H_{F}$ is a product of a certain number of the $G_{i}$ (Milne, Algebraic Groups, 21.51), which is stable under the Galois action. But $\Gal(F/\mathbb{Q})$ acts transitively on the set of $G_{i}$, and so $H=\Res_{F/\mathbb{Q}}G$.

In the general case, the same argument works if $G$ is geometrically simple. If not, then we may suppose that $G=\Res_{F^{\prime}/F}G^{\prime}$ with $G^{\prime }$ geometrically simple (ibid. 24.3) and apply the argument to $G^{\prime}$.

This argument fails for algebraic groups like $\mathbb{G}_a^n$ because for them the decomposition into simple objects is not unique.

As noted, the answer is yes. Let $G$ be a simple algebraic group (meaning $F$-simple) over $F$.

Assume first that $F$ is Galois over $\mathbb{Q}$. Then $(\mathrm{Res}_{F/\mathbb{Q}}G)_{F}\simeq G_{1}\times\cdots\times G_{r}$, where the $G_{i}$ are the conjugates of $G$ under $\mathrm{Gal}(F/\mathbb{Q})$. If $H$ is a smooth connected normal algebraic subgroup of $\mathrm{Res}_{F/\mathbb{Q}}G$, then $H_{F}$ is a product of a certain number of the $G_{i}$ (Milne, Algebraic Groups, 21.51), which is stable under the Galois action. But $\mathrm{Gal}% (F/\mathbb{Q})$ acts transitively on the set of $G_{i}$, and so $H=\mathrm{Res}_{F/\mathbb{Q}}G$.

In the general case, the same argument works if $G$ is geometrically simple. If not, then we may suppose that $G=\mathrm{Res}_{F^{\prime}/F}G^{\prime}$ with $G^{\prime}$ geometrically simple (ibid. 24.3) and apply the argument to $G^{\prime}$.

This argument fails for algebraic groups like $\mathbb{G}_{a}^{n}$ because for them the decomposition into simple objects is not unique.

Note that the proof only uses standard properties of semisimple groups, for which I gave a modern reference. I don't understand the comment of LSpice.

I posted a proof that the restriction is simple in the hope that it would be useful, but I'm not interested in playing games with LSpice, so I have deleted it.

$\DeclareMathOperator\Res{Res}\DeclareMathOperator\Gal{Gal}$As noted, the answer is yes. Let $G$ be a simple algebraic group (meaning $F$-simple) over $F$.

Assume first that $F$ is Galois over $\mathbb{Q}$. Then $(\Res_{F/\mathbb{Q}}G)_{F}\simeq G_{1}\times\cdots\times G_{r}$, where the $G_{i}$ are the conjugates of $G$ under $\Gal(F/\mathbb{Q})$. If $H$ is a smooth connected normal algebraic subgroup of $\Res_{F/\mathbb{Q}}G$, then $H_{F}$ is a product of a certain number of the $G_{i}$ (Milne, Algebraic Groups, 21.51), which is stable under the Galois action. But $\Gal(F/\mathbb{Q})$ acts transitively on the set of $G_{i}$, and so $H=\Res_{F/\mathbb{Q}}G$.

In the general case, the same argument works if $G$ is geometrically simple. If not, then we may suppose that $G=\Res_{F^{\prime}/F}G^{\prime}$ with $G^{\prime }$ geometrically simple (ibid. 24.3) and apply the argument to $G^{\prime}$.

This argument fails for algebraic groups like $\mathbb{G}_a^n$ because for them the decomposition into simple objects is not unique.

$\DeclareMathOperator\Res{Res}\DeclareMathOperator\Gal{Gal}$ As noted, the answer is yes. Let $G$ be a simple algebraic group (meaning $F$-simple) over $F$.

Assume first that $F$ is Galois over $\mathbb{Q}$. Then $(\Res_{F/\mathbb{Q}}G)_{F}\simeq G_{1}\times\cdots\times G_{r}$, where the $G_{i}$ are the conjugates of $G$ under $\Gal(F/\mathbb{Q})$. If $H$ is a smooth connected normal algebraic subgroup of $\Res_{F/\mathbb{Q}}G$, then $H_{F}$ is a product of a certain number of the $G_{i}$ (Milne, Algebraic Groups, 21.51), which is stable under the Galois action. But $\Gal(F/\mathbb{Q})$ acts transitively on the set of $G_{i}$, and so $H=\Res_{F/\mathbb{Q}}G$.

In the general case, the same argument works if $G$ is geometrically simple. If not, then we may suppose that $G=\Res_{F^{\prime}/F}G^{\prime}$ with $G^{\prime }$ geometrically simple (ibid. 24.3) and apply the argument to $G^{\prime}$.

This argument fails for algebraic groups like $\mathbb{G}_a^n$ because for them the decomposition into simple objects is not unique.

I posted this in the hope it would be useful. I'm not interested in playing games. Unless LSpice deletes his comment, I'll remove it.

I posted a proof that the restriction is simple in the hope that it would be useful, but I'm not interested in playing games with LSpice, so I have deleted it.