For typing convenience, define the symmetric matrix
$$\eqalign{
X &= W^TW-I \\
}$$
Write the nuclear norm in terms of this new variable.

Then find its differential and (sub)gradient wrt $W$.
$$\eqalign{
\lambda &= \|X\|_* \\
&= {\rm Tr}\big((X^TX)^{1/2}\big) \\
&= \pm{\rm Tr}(X)\quad\Big({\rm choose\,sign\,such\,that\;}\lambda>0\Big) \\
d\lambda
&= \pm{\rm Tr}(dX) \\
&= \pm{\rm Tr}(W^TdW+dW^TW) \\
&= \pm{\rm Tr}\Big((2W)^TdW\Big) \\
\frac{\partial \lambda}{\partial W}
&= \pm 2W =
\begin{cases}
+2W &{\rm if\;Tr}(W^TW-I)>0 \\
-2W &{\rm otherwise}
\end{cases} \\
}$$
If you're concerned about computational expense, this result is as *inexpensive* as one could reasonably hope to achieve.

### Update

The matrix sign function is defined such that
$$\eqalign{
S &= \operatorname{sign}(X) = X(X^2)^{-1/2} \\
I &= S^2 \quad\implies S^{-1} = S \\
XS &= SX \\
}$$

For a symmetric matrix
$${
(X^TX)^{1/2} = (X^2)^{1/2} = XS^{-1} = SX
}$$
Therefore $S$ can be used to write the nuclear norm and its gradient as
$$\eqalign{
\lambda &= \operatorname{Tr}(SX) \\
\frac{\partial \lambda}{\partial W} &= 2WS \\
}$$
The previous result is only valid when $S=\pm I,\,$ which is true for
some matrices.