OK, let me try again, maybe I'll get it right this time. I'll show that $P$ is positive definite. This will imply the claim because if $(P+iQ)(x+iy)=0$ with $x,y\in\mathbb R^n$, then $Px=Qy$, $Py=-Qx$, and by taking scalar products with $x$ and $y$, respectively, we see that $\langle x, Px \rangle = -\langle y, Py\rangle$, which implies that $x=y=0$. Here I use that $Q$ is symmetric.
Let me now show that $P>0$. Following math110's suggestion, we can simplify my original calculation as follows: Let $ B=B_n = P -\textrm{diag}(1,2,\ldots , n)$. For example, for $n=5$, this is the matrix $$ B_ 5= \begin{pmatrix} 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2 & 2\\ 1 & 2 & 3 & 3 & 3\\ 1 & 2 & 3 & 4 & 4\\ 1 & 2 & 3 & 4 & 5 \end{pmatrix} . $$ I can now (in general) subtract the $(n-1)$st row from the last row, then the $(n-2)$nd row from the $(n-1)$st row etc. This confirms that $\det B_n=1$. Moreover, the upper left $k\times k$ submatrices of $B_n$ are of the same type; they equal $B_k$. This shows that $B>0$, by Sylvester's criterion, and thus $P>0$ as well.