I'm going answer my own question, for I think I found a proof of the following theorem.
Theorem 1. Let $f:\mathbb{R}^k\times[a,b]\to\mathbb{R}$ be a function, and let $F:\mathbb{R}^k\to\mathbb{R}$ be defined by integrating out the last variable, i.e., $F(x_1,\dots,x_k)=\int_a^b f(x_1,\dots,x_k,t)\,\text{d}t$. If $f$ is of class $C^n$ for some $n\geq 1$, then so is $F$.
Proof.
Notice first that the function $F$ exists, since $f$ is at least of class $C^1$ and hence continuous in $t$, and hence integrable over $t$.
We'll use Theorem 18.9.1 from the book A Course in Mathematical Analysis (Part II) by D.J.H. Garling:
Theorem 2. (Garling.)
Suppose that $A$ is a compact Jordan measurable subset of $\mathbb R^k$, that $a
The theorem is much more general than we need, so we will use the following corollary, which follows immediately from the theorem. (Notice that we interchange the roles of $x$ and $t$.)
Corollary 3.
Suppose that $g:\mathbb R\times[a,b]\to\mathbb R$ is continuous and has continuous partial derivative $(\partial g/\partial x)(x,t)$. Set $G(x)=\int_a^b g(x,t)\text{d} t$. Then $G$ is differentiable on $\mathbb R$ and
\begin{align*}
\frac{\text{d} G}{\mathbb d x}(x)=\int_a^b\frac{\partial g}{\partial x}(x,t)\text{d} t
\end{align*}
We now have the ingredients to prove Theorem 1. We fix $x_1,\dots,x_{k-1}$, set $g(x_k,t) = f(x_1,\dots,x_k,t)$ and define
\begin{align*}
G(x_k)=\int_a^b g(x_k,t)\text{d} t = \int_a^b f(x_1,\dots,x_k,t) \text{d} t = F(x_1,\dots, x_k)
\end{align*}
Since $f$ is of class $C^k$, $f$ is continuous and $\partial g/\partial x_k$ is continuous. By the corollary, $G$ is differentiable with
\begin{align*}
\frac{\text{d} G}{\text{d} x_k}(x_k)=\int_a^b\frac{\partial g}{\partial x_k}(x_k,t)\text{d} t
\end{align*}
But by definition of the partial derivative this means precisely that $F$ admits a partial derivative with respect to $x_k$ and that
\begin{align*}
\frac{\partial F}{\partial x_k}(x_1,\dots,x_k)=\int_a^b\frac{\partial f}{\partial x_k}(x_1,\dots,x_k,t)\text{d} t
\end{align*}
Continuity follows from the following. Let $\epsilon>0$. If $U\subset\mathbb R^k$ is an open ball around $0$, then there is a compact set $K$ that contains $U$. Then $K\times[a,b]$ is compact and $\frac{\partial f}{\partial x_k}$ is uniformly continuous on this set. Hence there exists $\delta>0$ such that if $|(x_1',\dots,x_k')-(x_1,\dots,x_k)|<\delta$ for $(x_1',\dots,x_k')\in U$ and $(x_1,\dots,x_k)\in U$, then
\begin{align*}
\bigg|\frac{\partial f}{\partial x_k}(x_1',\dots x_k',t)-\frac{\partial f}{\partial x_k}(x_1,\dots,x_k,t)\bigg|<\epsilon/(b-a)
\end{align*}
But then
\begin{align*}
\bigg|\frac{\partial F}{\partial x_k}&(x_1',\dots x_k')-\frac{\partial F}{\partial x_k}(x_1,\dots,x_k)\bigg| \\
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&= \bigg|\int_a^b\frac{\partial f}{\partial x_k}(x_1',\dots x_k',t)\text{d} t-\int_a^b\frac{\partial f}{\partial x_k}(x_1,\dots,x_k,t)\text{d} t\bigg|\\
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&\leq \int_a^b\bigg|\frac{\partial f}{\partial x_k}(x_1',\dots x_k',t)-\frac{\partial f}{\partial x_k}(x_1,\dots,x_k,t)\bigg|\text{d} t \\
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&<\epsilon/(b-a)\int_a^b\text{d} t = \epsilon
\end{align*}
which shows that $\frac{\partial F}{\partial x_k}$ is continuous on $U$. But this holds for all open balls $U$ around $0$, and hence $\frac{\partial F}{\partial x_k}$ is continuous on $\mathbb R^k$.
Of course there is nothing special about $x_k$, and indeed, similar arguments show that $F$ admits a continuous partial derivative with respect to $x_i$ with
\begin{align*}
\frac{\partial F}{\partial x_i}(x_1,\dots,x_k)=\int_a^b\frac{\partial f}{\partial x_i}(x_1,\dots,x_k,t)\text{d} t
\end{align*}
for each $i=1,\dots, k$. Since all partial derivatives of $F$ are continuous, $F$ is of class $C^1$. If $n=1$ we are done. Provided $n\geq 2$, we repeat the exact same argument with $g(x_k,t) = \frac{\partial f}{\partial x_i}(x_1,\dots,x_k,t)$ (for arbitrary $i$) and
\begin{align*}
G(x_k)=\int_a^b g(x_k,t)\text{d} t = \int_a^b \frac{\partial f}{\partial x_i}(x_1,\dots,x_k,t) \text{d} t = \frac{\partial F}{\partial x_i}(x_1,\dots, x_k)
\end{align*}
The same argument is again possible because $g$ is continuous and $\partial g/\partial t$ is as well, as $f$ is now at least of class $C^2$. If we also realize again that $x_k$ is not special, it follows that $\partial F/\partial x_i$ admits a continuous partial derivative with respect to $x_j$, for all $j$, and that
\begin{align*}
\frac{\partial^2 F}{\partial x_i\partial x_j}(x_1,\dots,x_k)=\int_a^b\frac{\partial^2 f}{\partial x_i\partial x_j}(x_1,\dots,x_k,t)\text{d} t
\end{align*}
for all $i,j=1,\dots, k$. Hence $F$ is of class $C^2$. If $n=2$ we are done, and if $n>2$ we can continue in this way, repeating the same argument with the obvious choices for the function $g$. It thus follows that $F$ admits all partial derivatives of order $n$, and that these are all continuous, which means that $F$ is of class $C^n$, as was to be shown. Q.E.D.
Remark. The proof also shows that the partial derivatives of $F$ are obtained by `differentiating under the integral sign'.
