First, the notation is a little weird because it is in "operator" form. $\int \mathrm{d}x$ is the "antidifferentiate with respect to $x$ the expression to the right" with the usual semantics of definite integration when the integral sign is given bounds. An example of converting this operator notation to iterated integrals is (note the reversal of order of the differential elements):
\begin{align*}
\int_a^b \mathrm{d}x \int_c^d \mathrm{d}y F(x,y) &= \int_a^b \mathrm{d}x \left( \int_c^d \mathrm{d}y \left( F(x,y) \right) \right) \\
&=\int_a^b \; \int_c^d \; F(x,y) \,\mathrm{d}y \,\mathrm{d}x \text{.}
\end{align*}
(Students who have taken a class like Real Analysis, will see the reversal of order of the differential elements and be thinking something like "but, Fubini's theorem...". To which I reply: Maybe, depends on how badly behaved $F$ is.)
Compare with the operator form of the derivative, $\mathrm{d}/\mathrm{d}x$, which is "differentiate with respect to $x$ the expression to the right".
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \frac{\mathrm{d}}{\mathrm{d}y} F(x,y) &= \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\mathrm{d}}{\mathrm{d}y} \left( F(x,y) \right) \right) \\
&= \frac{\mathrm{d}\frac{\mathrm{d}F(x,y)}{\mathrm{d}y}}{\mathrm{d}x} \text{.}
\end{align*}
Remember that operator equations need to act on some function. This is similar to the functional equation $f = g-h$, which says, if you feed all the functions the same $x$ and evaluate, you'll get the value equation $f(x) = g(x) - h(x)$, and since this equation is true for all $x$s (in the domain of $f$ and hence also of $g$ and $h$), the functional equation holds. The operator equation says: if you feed the same function to the operators on the left and right sides of the equation and evaluate, you will get the same resulting function on both sides and since this is true for all functions in the domain of the operators, the operator equation holds.
The particular expression
$$ \int_0^V \mathrm{d}\vec{x}_1 \cdots \int_0^V \mathrm{d}\vec{x}_N $$
in your display equation for $Q$ very much looks like a cut-(edit a little)-paste error coming from an example involving a 1-spatial-dimension problem where the position vectors had one component in the interval $[0,V]$. I believe a better cut-(edit a little)-paste version would be
$$ \int_{\vec{x}_1 \in V} \mathrm{d}\vec{x}_1 \cdots \int_{\vec{x}_N \in V} \mathrm{d}\vec{x}_N \text{.} $$
The integrand is $\mathrm{e}^{-\beta U}$. Very, very conveniently, since there is no position-dependent potential, $U$ is independent of $\vec{x}_1$ (and the other $\vec{x}_i$), so the effect of the integral is to multiply the result by the measure of the set over which we integrate, $V$. Compare, in a setting that has $x$ and $y$ as independent variables,
\begin{align*}
\int_a^b \; y \,\mathrm{d}x &= \left. xy \right|_{x = a}^b \\
&= (b-a) y
\end{align*}
and, where $U$ is independent of $\vec{x}_1$,
\begin{align*}
\int_{\vec{x}_1 \in V} \mathrm{d}\vec{x}_1 \mathrm{e}^{-\beta U} &= V \mathrm{e}^{-\beta U} \text{.}
\end{align*}
(One could incorrectly write this as
\begin{align*}
\int_0^V \mathrm{d}\vec{x}_1 \mathrm{e}^{-\beta U} &= (V-0) \mathrm{e}^{-\beta U} \\
&= V \mathrm{e}^{-\beta U} \text{,}
\end{align*}
arriving at the right answer by doing the wrong thing.
)
The integral in the display for $Q$ has $N$ integrals as the $\vec{x}_i$ range over $V$, so those collapse to a premultiplier $V^N$ in front of the integrals over the momentum vectors. That is, because $U$ has no position dependent potential,
$$ Q = \frac{V^N}{N!} \int_{-\infty}^{\infty}\; \mathrm{d}\vec{p}_1 \cdots \int_{-\infty}^{\infty}\; \mathrm{d}\vec{p}_N \mathrm{e}^{-\beta U} \text{.} $$
If you'd like to see this done with slightly different notation, try https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Statistical_Mechanics/Boltzmann_Average/Ideal_Gas_Partition_Function There, equation (7) corresponds to the ideal gas and you can see the $V^N/N!$ in front of the term that is produced by doing all the (Gaussian) momentum integrals.
Finally, you ask about
$$ \int_{-\infty}^{\infty} \,\mathrm{d}\vec{p} = \int_{-\infty}^{\infty} \,\mathrm{d}p_x \int_{-\infty}^{\infty} \,\mathrm{d}p_y \int_{-\infty}^{\infty} \,\mathrm{d}p_z \text{.} $$
Remember this is an operator equation. It means that if you apply the operator on the left to a function you get the same thing as the operator on the right. The operator on the left looks weird for two reasons. First, it should be
$$ \int_{(-\infty,\infty)^3} \,\mathrm{d}\vec{p} \text{,} $$
the integral of a density over all of a 3-dimensional momentum space, and second, our integral is defined as a limit of Riemann sums as we partition that momentum space in the $\vec{p}$ direction. What that actually means is we should partition it into little 3-dimensional boxes and a smart way to do that is to
- partition the whole space into slabs perpendicular to the $x$-axis of this momentum space, $p_x$, preparing to integrate the slabs along the $x$-direction for the first operator on the right,
- partition each slab into noodles perpendicular to the $x$- and $y$-axes, preparing for the middle operator on the right, and finally
- partition each noodle along the $z$-direction into boxes, preparing for the last operator on the right.
So this is the integration version that most matches the idea you were thinking about in your angle-bracketed display.
(Note: The partition into noodles in each slab can be different from the partition into noodles in any of the other slabs. Similarly, the partitions into boxes can be unrelated between noodles. If you're imagining a partition into boxes made by planes parallel to the coordinate planes that completely cut through $V$, this is a valid partition scheme, but Riemann sums allow a lot more freedom that you almost never need and never want to think about, if you can avoid it.)
Note that if $V$ were a rectangular parallelpiped, $[a,b]\times[c,d]\times[e,f]$, the following operator equation holds:
$$ \int_{\vec{v} \in V} \,\mathrm{d}\vec{v} = \int_a^b \,\mathrm{d}\vec{v}_x \int_c^d \,\mathrm{d}\vec{v}_y \int_e^f \,\mathrm{d}\vec{v}_z $$
where $\vec{v}_x$ points along the $x$-axis and similarly for $\vec{v}_y$ and $\vec{v}_z$. This is probably the best way to get at what you were trying to express in the angle brackets. Notice that like this, we can't have $V$ be any other shape. Since the $\displaystyle \int_{\vec{v} \in V} \,\mathrm{d}\vec{v}$ operator hides all the details about the shape of $V$, it allows you to more easily see the physics. Actually evaluating the integral when applying that operator generally in problems requires writing it as a triple integral and working out all the expressions for the intervals of integration for each nested integral, so is better to think of as the version on the right-hand side of the operator equation.
