Cross Product of two Hermitian Operators The operator for linear momentum $\mathbf{p}$ and the operator for orbital angular momentum $\mathbf{L}$ ($\mathbf{L} = \mathbf{r} \times \mathbf{p}$) are Hermitian. Is the cross product between $\mathbf{p}$ and $\mathbf{L}$ also Hermitian? If I go component wise and use the commutator bracket relation $[p_i,L_j]=i\hbar \epsilon _{ijm}p_m$, then I get
$$(\mathbf{p} \times \mathbf{L})_x = p_yL_z - p_zL_y.$$
If I take the hermitian conjugate, I get
$$(\mathbf{p} \times \mathbf{L})^\dagger_x = L_zp_y - L_yp_z.$$
Now, $[L_i, p_j]$ doesn't commute if $i\neq j$. So, we get that $(\mathbf{p} \times \mathbf{L})_i \neq (\mathbf{p} \times \mathbf{L})^\dagger_i$ in this case. Thus, it's not Hermitian.
But on using the vector triple product relation, I see that it's Hermitian indeed. Can someone explain to me where am I going wrong?
Also, can we state this in general for any such cross product between two Hermitian operators?
 A: I'm not sure what triple product you have in mind. (I sincerely hope you are not conflicted about $([A,B])^\dagger = - [A,B]$  for any two hermitean operators A and B. This is the very reason physicists stick the i on the right-hand side of commutation relations!)
The standard procedure in physics, e.g. Pauli's legendary operator solution of the Hydrogen atom, is to hermitianize your operator by hand, e.g. in the celebrated LRL vector hermitianized to
$$
{\bf A}=({\bf p×L−L×p})/2-mk\hat {\bf r},
$$
which generates extra quantum shift terms, such as in
$$
{\bf A}^2= ({\bf p}^2 -2mk/r )~ ({\bf L}^2+\hbar^2) +m^2k^2 .
$$
($k= Ze^2$ is the strength of the Kepler/Coulomb potential.)
A: 
Is the cross product between p and L also Hermitian?

No it is not.

But on using the vector triple product relation, I see that it's Hermitian indeed.

No, even using a triple product relation, it's still not Hermitian.

Can someone explain to me where am I going wrong?

Nothing's going wrong with your cross product. You have used the wrong formula for your triple product, but even if you use the correct formula, the result is still not Hermitian (it's always anti-Hermitian).

As illustrative simpler cross-product example, even if you use the same Hermitian operator, the cross product does not have to be Hermitian:
$$
\vec L \times \vec L = i\vec L
$$
Even though $\vec L$ is Hermitian, $\vec L \times \vec L$ is anti-Hermitian.

And to follow up on the comment from OP on his post:

But if I write the vector product p×(r×p), I would get (p⋅p)r−(p⋅r)p.

Oh, but would you??
Actually, no, you would find:
$$
\vec p \times (\vec r \times\vec p)
$$
$$ 
= (\vec p \cdot \vec p) \vec r - (\vec p \cdot \vec r)\vec p + i\hbar \vec p
$$
$$
= (\vec p \cdot \vec p) \vec r - (\vec r \cdot \vec p)\vec p - 2i\hbar \vec p\;,
$$ (Or various other expressions depending on how you want to order the rs and the ps.)
But how you arrive at this result is a bit subtle (evidenced by the fact that I have screwed up this derivation about three times while trying to write up this answer too quickly).
First, using $\epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$, you find:
$$
\vec p \times (\vec r \times \vec p)
=p_k \vec r p_k - (\vec p \cdot \vec r) \vec p\;,
$$
where the sum over $k$ is implied.
But, since $\vec r$ and $\vec p$ are operators, you can't just commute the rs and ps. Well, you can commute them, but you need to properly account for the factors of $i\hbar$:
$$
p_k \vec r p_k - (\vec p \cdot \vec r) \vec p
$$
$$
=p^2\vec r +i\hbar\vec p - (\vec p \cdot \vec r)\vec p
$$
$$
=p^2\vec r +i\hbar\vec p - (\vec r \cdot \vec p - 3i\hbar)\vec p
$$
$$
=p^2\vec r - (\vec r \cdot \vec p) \vec p - 2i\hbar \vec p\;.
$$
