I also don't know what a Hermitian surface is, but it seems to me that there are lots of these surfaces for any given set of lines.

Let $L$ be the union of $m$ pairwise disjoint union of copies of $\mathbb P^r$ in $\mathbb P^n$ over a field $k$ and let $\mathscr I\subseteq \mathscr O_{\mathbb P^n}$ denote its ideal sheaf. A degree $d$ hypersurface in $\mathbb P^n$ containing $L$ corresponds to a global section of $\mathscr I(d)\subseteq \mathscr O_{\mathbb P^n}(d)$. So, (it seems) all you need is that $d$ is large enough so this would not be zero. This will follow as soon as
$$
\dim_k H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(d)) >
\dim_k \oplus^mH^0(\mathbb P^r, \mathscr O_{\mathbb P^r}(d)),
$$
i.e., when
$$
{n+d \choose d} > m\cdot{r+d \choose d}.
$$
In particular, if $L$ is a disjoint union of $3$ lines in $\mathbb P^3$, then one needs that
$$
{d+3\choose d} > 3 (d+1).
$$
It seems to me that this happens as soon as $d\geq 2$. If most surfaces are not Hermitian, then this should give you plenty of examples as $d$ grows.

Note that this computation does not use that these are lines, only that they are isomorphic to $\mathbb P^1$ (or $\mathbb P^r$ in the general case), so you could include other rational curves. Furthermore, one can easily develop a formula for arbitrary curves where the actual bound on the degree would depend on the genus of the curves involved, but in any case I would expect that for any fixed genus one would get surfaces containing a fixed set of curves if one allows the degree to be sufficiently large.

**EDIT:** As *M.D.* correctly points out this procedure might produce reducible hypersurfaces. However, using the same formula one can count those and conclude that, at least for large $d$, there is still enough dimension left so the general member is indeed irreducible.

For simplicity let's do this for $n=3$, but the method is the same in higher dimensions: A reducible degree $d$ surface is a union of smaller degree surfaces. We can just think of a union of two who themselves may be reducible. So how many of those are there?
Let $L_{i,d-i}\subseteq H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(4))$ denote the images of the product maps
$$
H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(i))\times H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(6-i)) \longrightarrow H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(6))
$$
for $i=1,\dots,\left[d/2\right]$.

Since (as above) $\dim_k H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(j))={j+3\choose 3}$, we get that

$\dim_kL_{i,d-i}\leq {i+3\choose 3} + {d-i+3\choose 3}$

At the same time, according to the above argument, the subspace of $H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(d))$ containing three skew lines is at least
$$
{d+3\choose 3} - 3 (d+1),
$$
so as soon as
$$
{d+3\choose 3} - 3 (d+1) - \max_{0<i<d}\left({i+3\choose 3} + {d-i+3\choose 3}\right) >0, \tag{$\star$}
$$
the general such surface will be irreducible. It is easy to see that the left hand side of $(\star)$ is a quadratic polynomial with positive leading coefficient, so this will be positive for $d\gg 0$. In fact, I think it is positive for $d\geq 6$.

Note that this was a very rough calculation as we counted all reducible surfaces, not just the ones containing the three lines, so one can probably do better for smaller degrees, but I leave that to the reader.