Let $u\in \mathbb{Z}_p^*$ be a unit in the ring of $p$-adic integers. Assume that $u^{1/p}\not\in \mathbb{Q}_p$, in other words $u$ is not a $p$-power.
Define by $K:=\mathbb{Q}_p(u^{1/p})$. What is the discriminant of $K$?
The discriminant of the defining polynomial $f(x):=x^p-u$ is $\Delta(f)=(-1)^{p(p-1)/2}p^pu^{p-1}$, and I know that $\Delta(K/\mathbb{Q}_p)|\Delta(f)$, however can one explicitly compute $\Delta(K/\mathbb{Q}_p)$?
I think the question reduces to how do I find an integral basis for $\mathcal{O}_K$, or how do I find a uniformizer for $\mathcal{O}_K$?
You'll need to consider whether or not $u^{p-1} \equiv 1 \bmod p^2$. This "Wieferich prime" criterion comes up in the number field setting in order to determine whether $\mathbf Q(\sqrt[p]{a})$ has ring of integers $\mathbf Z[\sqrt[p]{a}]$ when $p$ is prime and $a$ is a squarefree integer other than $\pm 1$: see the last theorem here. It is also relevant in the local case.
You tell us that $u$ is not a $p$th power in $\mathbf Q_p$. That implies $x^p - u$ is irreducible over $\mathbf Q_p$ (over all fields, even those of characteristic $p$, $x^p - a$ is irreducible when $a$ is not a $p$th power in the field). Let $K = \mathbf Q_p(\sqrt[p]{u})$, so $[K:\mathbf Q_p] = p$. Since "$n=ef$" in local fields, when $n = p$ either $e = p$ or $f = p$. So $K$ is either totally ramified or unramified.
Theorem. If $u^{p-1} \not\equiv 1 \bmod p^2$ then $K/\mathbf Q_p$ is totally ramified with prime element $\sqrt[p]{u}-u$ and ring of integers $\mathbf Z_p[\sqrt[p]{u}]$.
Proof. The polynomial $(x+u)^p - u$, with root $\sqrt[p]{u}-u$, is Eisenstein at $p$ since its constant term $u^p - u$ is divisible by $p$ but not by $p^2$ due to the hypothesis in the theorem. (The intermediate coefficients are all obviously divisible by $p$.) Hence $K$, which is $\mathbf Q_p(\sqrt[p]{u}-u)$, is totally ramified with prime element $\sqrt[p]{u}-u$. In every totally ramified extension of $\mathbf Q_p$, the powers of an arbitrary prime in that extension is a power basis for its ring of integers, so the integers of $K$ equal $\mathbf Z_p[\sqrt[p]{u}-u]$, which is also $\mathbf Z_p[\sqrt[p]{u}]$. QED
Corollary. If $u^{p-1} \not\equiv 1 \bmod p^2$, then ${\rm disc}(K/\mathbf Z_p)$ equals ${\rm disc}(x^p - u)$ up to unit factors.
Proof. The discriminant equals ${\rm disc}(x^p - u)$ since the ring of integers has powers of $\sqrt[p]{u}$ as a basis. QED
Theorem. If $u^{p-1} \equiv 1 \bmod p^2$ then $K/\mathbf Q_p$ is unramified.
Proof. This can be seen by a direct calculation when $p = 2$: when $u \equiv 1 \bmod 4$ and $u$ is not a square, then $u \equiv 5 \equiv -3 \bmod 8$, so $\mathbf Q_2(\sqrt{u}) = \mathbf Q_2(\sqrt{-3})$, which is unramified over $\mathbf Q_2$.
Now we can suppose $p > 2$. We'll prove the theorem by proving the contrapositive: if $K/\mathbf Q_p$ is ramified then $u^{p-1} \not\equiv 1 \bmod p^2$.
Let $\alpha = \sqrt[p]{u}-u$, so $\alpha \not\in \mathbf Q_p$ and thus $K = \mathbf Q_p(\alpha)$, as $[K:\mathbf Q_p] = p$, a prime. Thus the minimal polynomial of $\alpha$ over $\mathbf Q_p$ has to be of degree $p$. Since $\alpha$ is obviously a root of $$ (x+u)^p - u $$ in $\mathbf Q_p[x]$, which has degree $p$, this is the minimal polynomial of $\alpha$ over $\mathbf Q_p$. Therefore ${\rm N}_{K/\mathbf Q_p}(\alpha) = \pm(u^p-u)$, which is $0 \bmod p$, so $|\alpha|_p = \sqrt[p]{|{\rm N}_{K/\mathbf Q_p}(\alpha)|_p} < 1$.
Suppose $K/\mathbf Q_p$ is ramified. Then, because $[K:\mathbf Q_p] = p$, the extension is totally ramified. Writing $\pi$ for a prime in $K$, $p$ and $\pi^p$ are unit multiples in $K$. Thus their norms down to $\mathbf Q_p$ are unit multiples in $\mathbf Q_p$, so $|{\rm N}_{K/\mathbf Q_p}(\pi)|_p = |p|_p$.
Let $r = {\rm ord}_{\pi}(\alpha)$, so $r \geq 1$ since $|\alpha|_p < 1$. Write $\alpha = \pi^r v$ for some unit $v$ in $K$. Taking norms of both sides and recalling that ${\rm N}_{K/\mathbf Q_p}(\alpha) = \pm(u^p-u)$, $$ |u^p - u|_p = |{\rm N}_{K/\mathbf Q_p}(\pi)|_p^r = |p|_p^r = |p^r|_p, $$ so ${\rm ord}_p(u^p-u) = r$. We are going to show $r = 1$, so the highest power of $p$ dividing $u^p - u$ is $p$. Thus the highest power of $p$ dividing $u^{p-1} - 1$ is also $p$, which implies $u^{p-1} \not\equiv 1 \bmod p^2$.
Since $$ 0 = (\alpha + u)^p - u = \alpha^p + \sum_{i=1}^{p-1} \binom{p}{i}\alpha^i u^{p-i} + (u^p-u), $$ we have $$ \alpha^p + \sum_{i=1}^{p-1} \binom{p}{i}\alpha^i u^{p-i} = -(u^p-u). $$ We'll determine the $\pi$-adic valuation of both sides.
Since ${\rm ord}_{\pi}(\alpha) = r$ and ${\rm ord}_{\pi}(p) = p$, on the left side we have ${\rm ord}_{\pi}(\alpha^p) = pr$ and the $i$th term in the sum has $\pi$-adic valuation $p + ri$. Among the valuations $pr, p + r, \ldots, p + (p-1)r$, the unique minimal value is $p+r$ since $p+r < pr$ due to $p$ being odd and $r \geq 2$. (We can have $p+r = pr$ only when $p=2$ and $r = 2$, and $pr < p+r$ when $r = 1$). That makes the left side above have $\pi$-adic valuation $p+r$.
On the right side above, ${\rm ord}_p(u^p-u) = r$, so ${\rm ord}_{\pi}(u^p-u) = pr$. Thus $p+r = pr$, which is impossible when $p > 2$ and $r \geq 2$.
This contradiction shows $r = 1$ when $p > 2$. We already saw that having $r = 1$ makes $u^{p-1} \not\equiv 1 \bmod p^2$. QED
Corollary. If $u^{p-1} \equiv 1 \bmod p^2$, then ${\rm disc}(K/\mathbf Z_p)$ equals $1$ up to unit factors.
Proof. The discriminant is divisible by $p$ if and only if the extension is ramified, and $K/\mathbf Q_p$ is unramified in this case. QED
When $u^{p-1} \equiv 1 \bmod p^2$, the ring of integers of $K$ is bigger than $\mathbf Z_p[\sqrt[p]{u}]$ since the discriminant of $\mathbf Z_p[\sqrt[p]{u}]$ over $\mathbf Z_p$ is ${\rm disc}(x^p- u)$, which is divisible by $p$.