I also can't answer the question, but I'll say some things that could help. One thing von Staudt-Clausen tells you is the denominator of the Bernoulli number $B_k$ when p-1|k: it is precisely, the product of primes p for which p-1|k (when $p-1\nmid k$, a result of Kummer says that $B_k/k$ is p-integral). As Buzzard commented, the Bernoulli numbers should be thought of (at least in this situation) as appearing in special values of p-adic L-functions, specifically, for k a positive integer $$\zeta_p(1-k)=(1-p^{k-1})(-B_k/b),$$ where $\zeta_p$ is the p-adic Riemann zeta function (see chapter II of Koblitz's "p-adic numbers, p-adic analysis, and zeta-functions", for example). For a totally real field F, the Deligne-Ribet p-adic L-function provides a generalization of the p-adic Riemann zeta function, namely the p-adic Dedekind zeta function $\zeta_{F,p}$. One link between these and the Leopoldt conjecture is through the p-adic analytic class number formula which is the main theorem of Colmez's "Résidue en s=1 des fonctions zêta p-adiques": $$\lim_{s\rightarrow1}(s-1)\zeta_{F,p}(s)=\frac{2^{[F:\mathbf{Q}]}R_phE_p}{w\sqrt{D}}$$ where h is the class number, $E_p$ is a product of Euler factors, w is the number of roots of unity, D is the discriminant and $R_p$ is the interesting part here: the p-adic regulator.
Theorem: The Leopoldt conjecture is equivalent to the non-vanishing of the p-adic regulator.
A clear consequence of this is that if $\zeta_{F,p}$ does not have a pole at s=1, then the Leopoldt conjecture is false for (F,p). Perhaps an understanding of the denominators of values of $\zeta_{F,p}$ could lead to an understanding of the pole at s=1 of $\zeta_{F,p}$.