To flesh out Helge's answer a bit before I go to bed:

Assume that $f(x,y)$ is a smooth function on the unit square. 
Define the functions $f_n(x,y) = f(\frac{\lfloor nx \rfloor}{n}, \frac{\lfloor ny\rfloor}{n})$. This is a piecewise step function. Observe that the operator $S$ of dimension $d$ is the same if you define it relative to $f$ or $f_d$. It is elementary to show that $f_n\to f$ uniformly (as functions).

Define an action of $f_n$ on $L^2[0,1]$ by 
$$ g(x) \mapsto \int_0^1 f_n(x,y)g(y) dy $$
and note that for a $n$-vector $v = (v_1,\ldots, v_n)$ we can associate 
$$ g_v(x) = \sum v_i \chi_{[\frac{i-1}{n},\frac{i}{n})} $$ 
we observe that 
$$ f_n(g_v(x)) = \frac{1}{n} g_{Sv}(x)$$
the $1/n$ factor coming from the fact that the length of the segment $[(i-1) / n, i/n)$ is $1/n$. 

Now consider $V^n$ as the subspace of $L^2[0,1]$ spanned by $\chi_{[\frac{i-1}{n},\frac{i}{n})}$. By definition $f_n$ annihilates its orthogonal complement, and $f_n$ restricted to $V^n$ is equivalent to a rescaled version of $S$, so in particular you have that the trace norms of $f_n$ (acting on $L^2$) is the same as $1/n$ times the trace norm of $S$ (acting on $\mathbb{R}^n$). 

To finish you just need to note that via some functional analysis voodoo the corresponding operators $f_n\to f$, and so the trace norms of $f_n$ converges. Therefore you have that $1/n$ times the trace norm of $S$ converges to a constant (which may be zero). Note that the regularity for $f$ is only needed in this last step, and you probably just need  uniform continuity to assure that the operators converge in a strong enough sense.

[Addendum: all the functional analysis voodoo you need (which is not very much for this problem) can be found in Reed & Simon, volume 1, chapter 6.]