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This is a partial answer, only for $p=2$.

To simplify the writing, suppose that the dimension is $2$. Then, without loss of generality, $u$ is supported on $S:=[0,1]^2$. Let $D_j$ denote the operator of the partial differentiation wrt the $j$th argument. Then, for real $p\ge1$ and $(x,y)\in S$, by Jensen's inequality, $$|u(x,y)|^p\le\Big(\int_0^x (x-s)\,|(D_1^2u)(s,y)|\,ds\Big)^p \le\int_0^1|(D_1^2u)(s,y)|^p\,ds,$$ $$|D_1u(x,y)|^p\le\Big(\int_0^x|(D_1^2u)(s,y)|\,ds\Big)^p \le\int_0^1|(D_1^2u)(s,y)|^p\,ds,$$ so that $$\|u\|_p\le\|D_1^2u\|_p,$$ $$\|D_1u\|_p\le\|D_1^2u\|_p,$$ and similarly $$\|D_2u\|_p\le\|D_2^2u\|_p.$$

It remains to note that $$2\|D_1D_2u\|_2^2\le\|D_1^2u\|_2^2+\|D_2^2u\|_2^2,$$ which follows by using the Plancherel isometry for the Fourier transforms of the partial derivatives of $u$ of the second order.

It is noted in Willie Wong's answer that the definition of the operator $D^2$ in the OP is different from the one in the paper cited by the OP. Namely, in the paper $D^2$ involves both the mixed and unmixed partial derivatives, whereas in the OP $D^2$ involves only the unmixed partial derivatives.

However, the original question just as stated in the OP still makes sense, and answering that original question involves the additional task of bounding the $p$-norms of the mixed partial derivatives by the sum of the $p$-norms of the unmixed partial derivatives.

This was done in the previous version of this answer only for $p=2$, but now such a bound is available for all $p\in(1,\infty)$.

To simplify the writing, suppose that the dimension is $2$. Then, without loss of generality, $u$ is supported on $S:=[0,1]^2$. Let $D_j$ denote the operator of the partial differentiation wrt the $j$th argument. Then, for real $p\ge1$ and $(x,y)\in S$, by Jensen's inequality, $$|u(x,y)|^p\le\Big(\int_0^x (x-s)\,|(D_1^2u)(s,y)|\,ds\Big)^p \le\int_0^1|(D_1^2u)(s,y)|^p\,ds,$$ $$|D_1u(x,y)|^p\le\Big(\int_0^x|(D_1^2u)(s,y)|\,ds\Big)^p \le\int_0^1|(D_1^2u)(s,y)|^p\,ds,$$ so that $$\|u\|_p\le\|D_1^2u\|_p,$$ $$\|D_1u\|_p\le\|D_1^2u\|_p,$$ and similarly $$\|D_2u\|_p\le\|D_2^2u\|_p.$$

It remains to note that, thanks to this recent answer,
$$\|D_1D_2u\|_p\le C_p(\|D_1^2u\|_p+\|D_2^2u\|_p)$$ for each $p\in(1,\infty)$ and some real constant $C_p$ depending only on $p$.

This is a partial answer, only for $p=2$.

To simplify the writing, suppose that the dimension is $2$. Then, without loss of generality, $u$ is supported on $S:=[0,1]^2$. Let $D_j$ denote the operator of the partial differentiation wrt the $j$th argument. Then, for real $p\ge1$ and $(x,y)\in S$, by Jensen's inequality, $$|u(x,y)|^p\le\Big(\int_0^x |x-s||D_1^2f(s,y)|\,ds\Big)^p \le\int_0^1|D_1^2f(s,y)|^p\,ds,$$ $$|D_1u(x,y)|^p\le\Big(\int_0^x|D_1^2f(s,y)|\,ds\Big)^p \le\int_0^1|D_1^2f(s,y)|^p\,ds,$$ so that $$\|u\|_p\le\|D_1^2u\|_p,$$ $$\|D_1u\|_p\le\|D_1^2u\|_p,$$ and similarly $$\|D_2u\|_p\le\|D_2^2u\|_p.$$

It remains to note that $$2\|D_1D_2u\|_2^2\le\|D_1^2u\|_2^2+\|D_2^2u\|_2^2,$$ which follows by using the Plancherel isometry for the Fourier transform of the partial derivatives of $u$ of the second order.

This is a partial answer, only for $p=2$.

To simplify the writing, suppose that the dimension is $2$. Then, without loss of generality, $u$ is supported on $S:=[0,1]^2$. Let $D_j$ denote the operator of the partial differentiation wrt the $j$th argument. Then, for real $p\ge1$ and $(x,y)\in S$, by Jensen's inequality, $$|u(x,y)|^p\le\Big(\int_0^x (x-s)\,|(D_1^2u)(s,y)|\,ds\Big)^p \le\int_0^1|(D_1^2u)(s,y)|^p\,ds,$$ $$|D_1u(x,y)|^p\le\Big(\int_0^x|(D_1^2u)(s,y)|\,ds\Big)^p \le\int_0^1|(D_1^2u)(s,y)|^p\,ds,$$ so that $$\|u\|_p\le\|D_1^2u\|_p,$$ $$\|D_1u\|_p\le\|D_1^2u\|_p,$$ and similarly $$\|D_2u\|_p\le\|D_2^2u\|_p.$$

It remains to note that $$2\|D_1D_2u\|_2^2\le\|D_1^2u\|_2^2+\|D_2^2u\|_2^2,$$ which follows by using the Plancherel isometry for the Fourier transforms of the partial derivatives of $u$ of the second order.

This inequality is false. A counterexample: $u(x_1,x_2)=f(x_1)f(x_2)$, where $$f(x):=f_a(x):=x\,1(|x|<a) \\ +(a-|x|+1)^2 \left(-2 a^2+2 a |x|+|x|\right)\,1(a\le|x|<a+1)\,\text{sign}\, x$$ for $a\to\infty$ and real $x$.

The function $f$ is odd, in $C^1(\mathbb R)$, and supported on the interval $[-a-1,a+1]$. Moreover, $$\|u\|_2^2\asymp a^6,$$ whereas $$\|D^2u\|_2^2\asymp a^5$$ (as $a\to\infty$).

Below are the graphs of $f_{10}$ over the intervals $[-12,12]$ (left), $[9.9,10.1]$ (center), and $[10.9,11.1]$ (right):

This is a partial answer, only for $p=2$.

To simplify the writing, suppose that the dimension is $2$. Then, without loss of generality, $u$ is supported on $S:=[0,1]^2$. Let $D_j$ denote the operator of the partial differentiation wrt the $j$th argument. Then, for real $p\ge1$ and $(x,y)\in S$, by Jensen's inequality, $$|u(x,y)|^p\le\Big(\int_0^x |x-s||D_1^2f(s,y)|\,ds\Big)^p \le\int_0^1|D_1^2f(s,y)|^p\,ds,$$ $$|D_1u(x,y)|^p\le\Big(\int_0^x|D_1^2f(s,y)|\,ds\Big)^p \le\int_0^1|D_1^2f(s,y)|^p\,ds,$$ so that $$\|u\|_p\le\|D_1^2u\|_p,$$ $$\|D_1u\|_p\le\|D_1^2u\|_p,$$ and similarly $$\|D_2u\|_p\le\|D_2^2u\|_p.$$

It remains to note that $$2\|D_1D_2u\|_2^2\le\|D_1^2u\|_2^2+\|D_2^2u\|_2^2,$$ which follows by using the Plancherel isometry for the Fourier transform of the partial derivatives of $u$ of the second order.