This answer will be based om Fubini's theorem and DCT.
Put
\begin{equation*}
I_n = \int_{V_n}\dfrac{x_1^p+x_2^p+\dots +x_n^p}{x_1^q+x_2^q+\dots +x_n^q}\, dx_1dx_2\dots dx_n
\end{equation*}
where $ V_n=(0,1)^n. $ We will try to prove that
\begin{equation*}
\lim_{n\to \infty}I_n = \dfrac{q+1}{p+1} \tag{1}
\end{equation*}
if $ p > -1$ and $ q \ge 1$.
With $p=0$ and $q=1$ this will answer OP's question.
Before the proof we need some preparations.
Put
\begin{equation*}
c_q = \int_{0}^{\infty}e^{-x^q}\, dx = \dfrac{\Gamma\left(\dfrac{1}{q}\right)}{q}
\end{equation*}
and
\begin{equation*}
\mathrm{eq}=\dfrac{1}{c_q}\int_{x}^{\infty}
e^{-y^q}\, dy.
\end{equation*}
In order to later find a majorant we will use an inequality by $@$mickep (email communication in the case $q=1$).
\begin{equation*}
c_q\dfrac{1-\mathrm{eq}(t)}{t}\le \left(1+\dfrac{t^q}{2}\right)^{-\dfrac{1}{q}}, \quad t>0.\tag{2}
\end{equation*}
Proof of (2). Put
\begin{equation*}
f(t) = \dfrac{t}{\left(1+\dfrac{t^q}{2}\right)^{\dfrac{1}{q}}} - c_q(1-\mathrm{eq}(t)), \quad t \ge 0.
\end{equation*}
Then $f(0)=0$. We intend to prove that $ f'(t) \ge 0 $ if $ t \ge 0. $ We observe that
\begin{gather*}
f'(t) = \dfrac{1}{\left(1+\dfrac{t^q}{2}\right)^{\dfrac{1}{q}+1}} -e^{-t^q} \ge 0
\\
\Longleftrightarrow\\
e^{t^q} \ge \left(1+\dfrac{t^q}{2}\right)^{\dfrac{1}{q}+1} \\
\Longleftrightarrow\\
t^{q} \ge \left(\dfrac{1}{q}+1\right)\ln\left(1+\dfrac{t^q}{2}\right)\\
\Longleftrightarrow\\
\dfrac{2q}{q+1}\dfrac{t^q}{2} \ge \ln\left(1+\dfrac{t^q}{2}\right)
\end{gather*}
Since $ x \ge \ln(1+x)$ and $\dfrac{2q}{q+1} \ge 1 $ this is
true and we have proved (2).
Proof of (1).
\begin{gather*}
I_n=[\mbox{ symmetry}] = n\int_{0}^{1}x_1^p\left(\int_{0}^{\infty}e^{-t(x_1^q+x_2^q+\dots +x_n^q)}\, dt\right)\, dx_1dx_2\dots dx_n =
\\[2ex] n\int_{0}^{1}\int_{0}^{\infty}x_1^pe^{-tx_1^q}\left(\int_{0}^{1}e^{-ty^q}\, dy\right)^{n-1}\, dx_1dt = \left[z=t^{1/q}y\right] =\\[2ex]n\int_{0}^{1}\int_{0}^{\infty}x_1^pe^{-tx_1^q}\left(\int_{0}^{t^{1/q}}\dfrac{e^{-z^q}}{t^{1/q}}\, dz\right)^{n-1}\, dx_1dt =\\[2ex]
n\int_{0}^{1}\int_{0}^{\infty}x_1^pe^{-tx_1^q}\left(\dfrac{1}{t^{1/q}}\left[-c_q\mathrm{eq}(z)\right]_{0}^{t^{1/q}}\right)^{n-1}\, dx_1dt = \left[t=\frac{s}{n-1}\right]=\\[2ex] \dfrac{n}{n-1 }\int_{0}^{1}\int_{0}^{\infty}x_1^pe^{-sx_1^q/(n-1)}\left(\dfrac{1}{s_{qn}}c_q(1-\mathrm{eq}(s_{qn}))\right)^{n-1}\, dx_1ds\tag{3}
\end{gather*}
where $ s_{qn}= \left(\dfrac{s}{n-1}\right)^{1/q} $.
Now we will use Mickep's inequality (2) to find a majorant.
\begin{gather*}
e^{-sx_1^p/(n-1)}\left(\dfrac{1}{s_{qn}}c_q(1-\mathrm{eq}(s_{qn}))\right)^{n-1} \le 1 \cdot \left(\left(1+\dfrac{s}{2(n-1)}\right)^{-1/q}\right)^{n-1} =\\[2ex]
\left(\left(1+\dfrac{s}{2(n-1)}\right)^{n-1}\right)^{-1/q} \le \left(\left(1+\dfrac{s}{2(N-1)}\right)^{N-1}\right)^{-1/q} , \quad n \ge N
\end{gather*}
In the last inequality we have used that
\begin{equation*}
\left(1+\dfrac{s}{2(n-1)}\right)^{n-1}
\end{equation*}
is increasing towards $ e^{s/2} $. If we choose $ N $ such that $ \dfrac{N-1}{q}>1 $ the majorant
\begin{equation*}
\left(\left(1+\dfrac{s}{2(N-1)}\right)^{N-1}\right)^{-1/q}
\end{equation*}
will belong to $L_1.$
Finally we will study the pointwise limit.
Put
\begin{equation*}
g(x) = c_q(1-\mathrm{eq}(x)).
\end{equation*}
Then
\begin{equation*}
g'(x) = e^{-x^{q}} = 1- x^q + x^{2q}\cdot B_{1}(x^q)
\end{equation*}
where $B_1$ is q bounded function in the neighbourhood of origin. We get that
\begin{equation*}
g(x) = x-\dfrac{x^{q+1}}{q+1}+x^{2q+1}B_2 \tag{4}
\end{equation*}
where $B_{2}$ is bounded for small $x^{q}$. However, from (4) we get
\begin{gather*}
\left(\dfrac{1}{s_{qn}}c_q(1-\mathrm{eq}(s_{qn}))\right)^{n-1}=\left(1-\dfrac{s}{(q+1)(n-1)}+\dfrac{s^2}{(n-1)^{2}}B_3\right)^{n-1} \to e^{-s/(q+1)},\quad n \to \infty
\end{gather*}
since $B_{3}$ is bounded.
Now we return to (3).
\begin{equation*}
\lim_{n\to \infty}I_n = \int_{0}^{1}\int_{0}^{\infty}x_1^pe^{-s/(q+1)}\, dx_1ds = \dfrac{q+1}{p+1}.
\end{equation*}