Let's work locally at the prime $\mathfrak p$ (or complete at $\mathfrak p$ if you prefer), and so assume $\mathcal O_K$ is a DVR with maximal ideal $\mathfrak p$. Also, let's assume that $\mathcal O_L = \mathcal O_K[\alpha]$ for some $\alpha$. (Even if we pass to the completion at $\mathfrak p$ this isn't always true in the inseparable residue field context, because it's not always true that inseparable field extensions are simple, but it makes things more transparent if we assume this.)
If $f(x) \in \mathcal O_K[x]$ is the minimal polynomial of $\alpha$,
then we can write $\mathcal O_L = \mathcal O_K[x]/(f(x)) .$
Now unramified means that this extension should be etale in the sense of commutative algebra/algebraic geometry, i.e. that the derivative $f'(x)$ should be invertible in the quotient $\mathcal O_L$. (Actually etale equals unramified plus flat, but flatness of $\mathcal O_L$ over $\mathcal O_K$ is automatic in our setting; so don't worry if this doesn't mean anything to you!) Now being invertible in a DVR just means that $f'(x)$ is non-zero in the residue field, i.e. that $k[X]/(f(X))$ is a product of separable field extensions if, where $k = \mathcal O_K/\mathfrak p$, i.e. that $\mathcal O_L/\mathfrak p$ is a product of separable field extensions of $k$. In other words, $\mathcal O_L$ is unramified over $\mathcal O_K$ precisely if $\mathfrak p$ factors into a product of primes in $\mathcal O_L$, each appearing with multiplicity one, and each having a residue field that is separable over $k$.
For a geometric perspective, note that a non-separable residue field extension in the context of curves over a field will turn into multiplcity if we extend scalars.
E.g. consider the field $k = \mathbb F_p(T)$, and the map $\phi: \mathbb A^1_{/k} \to \mathbb A^1_{/k} $ defined by $X \mapsto T^{-1}X^p.$ (Here $X$ is the coordinate on $\mathbb A^1$; and remember that $T$ is an element of the ground field $k$.) This gives the extension of Dedekind domains (and $k$-algebras)
$$\mathcal O_K :=k[Y] \subset k[Y][X]/(X^p - TY)= k[X] =: \mathcal O_L.$$
If we take $\mathfrak p = (Y)$, corresponding to the point $0$ in $\mathbb A^1$,
then it factors as $(Y) =(TY)$ (remember $T$ is a unit) $= (X)^p$, so the extension of residue fields is trivial (just $k$ over $k$), while $e = p$.
On the other hand, if we look over the point $(Y -1)$, corresponding to the points $1$ in $\mathbb A^1$, there is a unique prime lying over it, namely $(X^p - T)$. So $e = 1$, but the residue field extension is non-trivial, and non-separated (it is $K(T^{1/p})$ over $k$.)
Now let's extend sclaras from $k$ to the algebraic closure of $k$ (or even just to $l := k(T^{1/p})$). Then our extension of Dedekind domains becomes
$$l[Y] \subset l[Y][X]/(X^p - TY) = l[Y][X] / (X - T^{1/p}Y)^p = l[X].$$ Above $0$ the fibre doesn't change, we still have $e = p$ and a trivial residue field extensions, but above $1$, we now have $(Y-1) = (X - T^{1/p})^p$, i.e. $e = p$ and a trivial extension of residue fields.
Our original map $\phi$ was not the Frobenius map, but was a twisted form of Frobenius (twisted by $T$). After extending scalars to $l$, we could untwist, and our map just becomes the Frobenius (after a change of coordinates to $T^{1/p}X$ on the source). So after this base-change, we have $e = p$ above every point/prime in the target; but before it, we didn't have $e = p$ in general --- rather we had $e = 1$ but an inseparable residue field extension of degree $p$.
So the notion of $e = 1$ is not stable under extension of scalars (when we are in the geometric context of curves over fields), but the notion of unramified ($e = 1$ and separable residue field extension) is stable under extension of scalars. Typically, properties that are separable under extension of scalars are better behaved --- of course, tautologically they behave better under extension of scalars (!), but this has a conceptual meaning: it means that they are capturing a true aspect of the underlying geometry, rather than some more transient phenomena that just has to do with us working over perhaps a field of scalars that is too small to reveal all the geometric phenomena in play.
Here I am restricting my discussion of extension of scalars to the setting of DVRs that come from curves over a field, because in the number theory setting (for instance) it is not so obvious how to make extensions of scalars that keep us inside the realm of standard algebraic number theory (DVRs inside finite extensions of $\mathbb Q$ or $\mathbb Q_p$). However, when defined properly, the concepts of unramified and etale make sense for any extension (or, more generally, morphism) of rings (or even more generally, morphism of schemes), and then both are stable under arbitrary base-change.
(In the formulation I gave above, writing $\mathcal O_L := \mathcal O_K[x]/(f(x))$ with $f'(x)$ being a unit in $\mathcal O_L$, this perservation under arbitrary base-change $\mathcal O_K \to A$ is pretty clear.)
Assuming that $L/K$ is Galois, the prime $\mathfrak{p}$ splits as
$$\prod_{i=1}^{r} \mathfrak{P}^e_i$$
where $\mathfrak{P}$ has inertial degree $f$ and $erf = [L:K]$. Assuming that the extension is tamely ramified, the different $\mathcal{D}_{L/K}$ coming from the primes above $\mathfrak{p}$ is equal to $\prod_{i=1}^{r} \mathfrak{P}^{e-1}_i$, and the discriminant is the norm of the different from $L$ to $K$, which is
$$\prod_{i=1}^{r} \mathfrak{p}^{f(e-1)} = \mathfrak{p}^{{rf(e-1)}} = \mathfrak{p}^{m},$$
where $m = [L:K]\left(1 - \frac{1}{e} \right)$.
If you don't assume that $L/K$ is Galois, then "ramification degree $e$" doesn't really mean anything --- some primes above $\mathfrak{p}$ may be ramified and some not.
If you don't assume that $L/K$ is tamely ramified then the answer will depend on more than just the invariant $e$. For example, if $K = \mathbf{Q}$, then $L = \mathbf{Q}(\sqrt{-1})$ and $\mathbf{Q}(\sqrt{2})$ are both Galois with $e = 2$ but the power of $2$ dividing the discriminant is $2^2$ or $2^3$.
On the other hand, you can determine the exponent if you know the orders of the higher ramification groups. In particular, one will have
$$m = [L:K] \left(\sum_{n=0}^{\infty} \frac{|I_n| - 1}{|I_0|}\right).$$
Here $|I_0| = e$ is the inertia group, and $|I_1|$ is the wild inertia group whose order is the largest power of $p$ dividing $|I_0|$. Since $|I_n| = 1$ for sufficiently large $n$, this is a finite sum. The difference between $L = \mathbf{Q}(\sqrt{-1})$ and $\mathbf{Q}(\sqrt{2})$ is that $\mathbf{Z}/2\mathbf{Z} = I_0 = I_1 \ne I_2$ in the first case, and $\mathbf{Z}/2\mathbf{Z} = I_0 = I_1 = I_2 \ne I_3$ in the second.
Best Answer
The fact that $L/\mathbb{Q}$ is unramified away from primes dividing $D=\text{disc } \mathcal{O}_K$ is evident: $L$ is composition of different embeddings of $K$, each such embedding is unramified away from primes dividing $D$, so is their composition $L$.
Now we show that for $p\mid D$, $p$ has ramification index $2$ in $L$. Let $\alpha_i\in L$, $i=1,\cdots,5$ be roots of $f(X) = X^5-X+1$. By factoring $f$ modulo $p$, we see that there are exactly four distinct $\bar{\alpha}_i \in \bar{\mathbb{F}}_p$, say $\bar{\alpha}_1 = \bar{\alpha}_2$ and $\bar{\alpha}_1, \bar{\alpha}_3,\bar{\alpha}_4,\bar{\alpha}_5$ are distinct. Any inertia group above $p$ fixes $\alpha_3,\alpha_4,\alpha_5$, only non-trivial element for inertia group will be the swapping of $\alpha_1$ and $\alpha_2$. Therefore ramification index is $2$.
To compute the discriminant, you can use the discriminant formula for tame ramification. But a more elegant approach is to consider $F = \mathbb{Q}(\sqrt{D})$. Since every $p\mid D$ has ramification $2$ in $L$, $L/F$ is unramified at every finite prime. Note that $[L:F] = 60$, therefore $$|D_{L/\mathbb{Q}}| = |D_{F/\mathbb{Q}}|^{60} = 19^{60} 151^{60}$$