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$\newcommand\E{E}\DeclareMathOperator\Var{Var}\newcommand\P{P}$Note that
$$\int_{[0,1)^d}\|x\|_2^p\,dx =\E S_d^{p/2},\tag{1}\label{1}$$ where $S_d:=\sum_1^d U_j^2$ and the $U_j$'s are iid random variables uniformly distributed on the interval $[0,1]$.

Note next that $\E S_d=d/3$ and $\Var S_d=4d/45<d/10$. So, by Cantelli's inequality, $$\P(S_d\ge d/6)\ge1-\frac{\Var S_d}{\Var S_d+(d/3-d/6)^2} \\ \ge 1-\frac{d/10}{d/10+(1/3-1/6)^2 d} \\ =1-\frac{1/10}{1/10+(1/3-1/6)^2}=:c\in(0,1).$$ So, $$\E S_d^{p/2}\ge(d/6)^{p/2} \P(S_d\ge d/6) \ge c(d/6)^{p/2}$$ and hence $$\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p} =(\E S_d^{p/2})^{1/p} \ge c^{1/p}\sqrt{d/6} \ge c\sqrt{d/6}$$ for $p\ge1$.

So, the trivial upper bound $\sqrt d$ on $\bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\bigr)^{1/p}$ is optimal up to a universal constant factor.

For $p\ge2$, one can can do with a much simpler reasoning: by Jensen's inequality, $$\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p} \ge\Bigl(\int_{[0,1)^d}\|x\|_2^2\,dx\Bigr)^{1/2} =\sqrt{d/3}.$$

One may also note that, for any real $p>0$, by the Fatou lemma, $$\liminf_{d\to\infty}\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p} \Big/\sqrt{d/3}\ge1.$$ In view of \eqref{1}, this follows because, by the law of large numbers, $S_d/d\to \E U_1^2=1/3$ in probability (as $d\to\infty$).

$\DeclareMathOperator\E{E}\DeclareMathOperator\Var{Var}\DeclareMathOperator\P{P}$Note that
$$\int_{[0,1)^d}\|x\|_2^p\,dx =\E S_d^{p/2},\tag{1}\label{1}$$ where $S_d:=\sum_1^d U_j^2$ and the $U_j$'s are iid random variables uniformly distributed on the interval $[0,1]$.

Note next that $\E S_d=d/3$ and $\Var S_d=4d/45<d/10$. So, by Cantelli's inequality, $$\P(S_d\ge d/6)\ge1-\frac{\Var S_d}{\Var S_d+(d/3-d/6)^2} \\ \ge 1-\frac{d/10}{d/10+(1/3-1/6)^2 d} \\ =1-\frac{1/10}{1/10+(1/3-1/6)^2}=:c\in(0,1).$$ So, $$\E S_d^{p/2}\ge(d/6)^{p/2} \P(S_d\ge d/6) \ge c(d/6)^{p/2}$$ and hence $$\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p} =(\E S_d^{p/2})^{1/p} \ge c^{1/p}\sqrt{d/6} \ge c\sqrt{d/6}$$ for $p\ge1$.

So, the trivial upper bound $\sqrt d$ on $\bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\bigr)^{1/p}$ is optimal up to a universal constant factor.

For $p\ge2$, one can can do with a much simpler reasoning: by Jensen's inequality, $$\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p} \ge\Bigl(\int_{[0,1)^d}\|x\|_2^2\,dx\Bigr)^{1/2} =\sqrt{d/3}.$$

One may also note that, for any real $p>0$, by the Fatou lemma, $$\liminf_{d\to\infty}\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p} \Big/\sqrt{d/3}\ge1.$$ In view of \eqref{1}, this follows because, by the law of large numbers, $S_d/d\to \E U_1^2=1/3$ in probability (as $d\to\infty$).

Note that
$$\int_{[0,1)^d}\|x\|_2^p\,dx =ES_d^{p/2},\tag{1}$$ where $S_d:=\sum_1^d U_j^2$ and the $U_j$'s are iid random variables uniformly distributed on the interval $[0,1]$.

Note next that $ES_d=d/3$ and $Var\,S_d=4d/45<d/10$. So, by Cantelli's inequality, $$P(S_d\ge d/6)\ge1-\frac{Var\,S_d}{Var\,S_d+(d/3-d/6)^2} \\ \ge 1-\frac{d/10}{d/10+(1/3-1/6)^2 d} \\ =1-\frac{1/10}{1/10+(1/3-1/6)^2}=:c\in(0,1).$$ So, $$ES_d^{p/2}\ge(d/6)^{p/2} P(S_d\ge d/6) \ge c(d/6)^{p/2}$$ and hence $$\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p} =(ES_d^{p/2})^{1/p} \ge c^{1/p}\sqrt{d/6} \ge c\sqrt{d/6}$$ for $p\ge1$.

So, the trivial upper bound $\sqrt d$ on $\big(\int_{[0,1)^d}\|x\|_2^p\,dx\big)^{1/p}$ is optimal up to a universal constant factor.

For $p\ge2$, one can can do with a much simpler reasoning: by Jensen's inequality, $$\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p} \ge\Big(\int_{[0,1)^d}\|x\|_2^2\,dx\Big)^{1/2} =\sqrt{d/3}.$$

One may also note that, for any real $p>0$, by the Fatou lemma, $$\liminf_{d\to\infty}\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p} \Big/\sqrt{d/3}\ge1.$$ In view of (1), this follows because, by the law of large numbers, $S_d/d\to EU_1^2=1/3$ in probability (as $d\to\infty$).

$\newcommand\E{E}\DeclareMathOperator\Var{Var}\newcommand\P{P}$Note that
$$\int_{[0,1)^d}\|x\|_2^p\,dx =\E S_d^{p/2},\tag{1}\label{1}$$ where $S_d:=\sum_1^d U_j^2$ and the $U_j$'s are iid random variables uniformly distributed on the interval $[0,1]$.

Note next that $\E S_d=d/3$ and $\Var S_d=4d/45<d/10$. So, by Cantelli's inequality, $$\P(S_d\ge d/6)\ge1-\frac{\Var S_d}{\Var S_d+(d/3-d/6)^2} \\ \ge 1-\frac{d/10}{d/10+(1/3-1/6)^2 d} \\ =1-\frac{1/10}{1/10+(1/3-1/6)^2}=:c\in(0,1).$$ So, $$\E S_d^{p/2}\ge(d/6)^{p/2} \P(S_d\ge d/6) \ge c(d/6)^{p/2}$$ and hence $$\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p} =(\E S_d^{p/2})^{1/p} \ge c^{1/p}\sqrt{d/6} \ge c\sqrt{d/6}$$ for $p\ge1$.

So, the trivial upper bound $\sqrt d$ on $\bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\bigr)^{1/p}$ is optimal up to a universal constant factor.

For $p\ge2$, one can can do with a much simpler reasoning: by Jensen's inequality, $$\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p} \ge\Bigl(\int_{[0,1)^d}\|x\|_2^2\,dx\Bigr)^{1/2} =\sqrt{d/3}.$$

One may also note that, for any real $p>0$, by the Fatou lemma, $$\liminf_{d\to\infty}\Bigl(\int_{[0,1)^d}\|x\|_2^p\,dx\Bigr)^{1/p} \Big/\sqrt{d/3}\ge1.$$ In view of \eqref{1}, this follows because, by the law of large numbers, $S_d/d\to \E U_1^2=1/3$ in probability (as $d\to\infty$).

Note that
$$\int_{[0,1)^d}\|x\|_2^p\,dx =ES_d^{p/2},$$ where $S_d:=\sum_1^d U_j^2$ and the $U_j$'s are iid random variables uniformly distributed on the interval $[0,1]$.

Note next that $ES_d=d/3$ and $Var\,S_d=4d/45<d/10$. So, by Cantelli's inequality, $$P(S_d\ge d/6)\ge1-\frac{Var\,S_d}{Var\,S_d+(d/3-d/6)^2} \\ \ge 1-\frac{d/10}{d/10+(1/3-1/6)^2 d} \\ =1-\frac{1/10}{1/10+(1/3-1/6)^2}=:c\in(0,1).$$ So, $$ES_d^{p/2}\ge(d/6)^{p/2} P(S_d\ge d/6) \ge c(d/6)^{p/2}$$ and hence $$\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p} =(ES_d^{p/2})^{1/p} \ge c^{1/p}\sqrt{d/6} \ge c\sqrt{d/6}$$ for $p\ge1$.

So, the trivial upper bound $\sqrt d$ on $\big(\int_{[0,1)^d}\|x\|_2^p\,dx\big)^{1/p}$ is optimal up to a universal constant factor.

For $p\ge2$, one can can do with a much simpler reasoning: by Jensen's inequality, $$\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p} \ge\Big(\int_{[0,1)^d}\|x\|_2^2\,dx\Big)^{1/2} =\sqrt{d/3}.$$

One may also note that, for any real $p>0$, by the Fatou lemma, $$\liminf_{d\to\infty}\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p} \Big/\sqrt{d/3}\ge1.$$

Note that
$$\int_{[0,1)^d}\|x\|_2^p\,dx =ES_d^{p/2},\tag{1}$$ where $S_d:=\sum_1^d U_j^2$ and the $U_j$'s are iid random variables uniformly distributed on the interval $[0,1]$.

Note next that $ES_d=d/3$ and $Var\,S_d=4d/45<d/10$. So, by Cantelli's inequality, $$P(S_d\ge d/6)\ge1-\frac{Var\,S_d}{Var\,S_d+(d/3-d/6)^2} \\ \ge 1-\frac{d/10}{d/10+(1/3-1/6)^2 d} \\ =1-\frac{1/10}{1/10+(1/3-1/6)^2}=:c\in(0,1).$$ So, $$ES_d^{p/2}\ge(d/6)^{p/2} P(S_d\ge d/6) \ge c(d/6)^{p/2}$$ and hence $$\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p} =(ES_d^{p/2})^{1/p} \ge c^{1/p}\sqrt{d/6} \ge c\sqrt{d/6}$$ for $p\ge1$.

So, the trivial upper bound $\sqrt d$ on $\big(\int_{[0,1)^d}\|x\|_2^p\,dx\big)^{1/p}$ is optimal up to a universal constant factor.

For $p\ge2$, one can can do with a much simpler reasoning: by Jensen's inequality, $$\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p} \ge\Big(\int_{[0,1)^d}\|x\|_2^2\,dx\Big)^{1/2} =\sqrt{d/3}.$$

One may also note that, for any real $p>0$, by the Fatou lemma, $$\liminf_{d\to\infty}\Big(\int_{[0,1)^d}\|x\|_2^p\,dx\Big)^{1/p} \Big/\sqrt{d/3}\ge1.$$ In view of (1), this follows because, by the law of large numbers, $S_d/d\to EU_1^2=1/3$ in probability (as $d\to\infty$).