Distributions are defined as the set of linear continuos operators over the set of test functions $ \mathcal{D}(\mathbb{R}) = C^{\infty}_c(\mathbb{R}) $, i.e. $ \mathcal{D}(\mathbb{R})^* $.
My question is:
are all distributions of the form
$$ \mathcal{D}(\mathbb{R}) \ni \varphi \mapsto \int_{\mathbb{R}} \varphi^{(n)}(x) d \mu (x) \in \mathbb{R} $$
(integral of differential operators against some measure) where $\varphi^{(n)}$ is the $n-$th derivative of $f$, and $\mu$ is a measure (or possibly linear combinations of such functionals)?
If the distribution $u$ has finite order (for instance, if it’s a tempered distribution) then yes: you can simply take $\mu=(-1)^n\partial_x^{-n}u$, recalling that taking the antiderivative of any distribution is always a well-defined operation (although the antiderivative is only uniquely defined up to an additive constant). For $n$ large enough, $\mu$ will actually be (say) a continuous function on all $\mathbb R$ if $u$ has finite order.
If $u$ has infinite order, for instance if $u:=\sum_{n=1}^\infty \partial_x^n\delta(x-n)$, then the above holds only locally in $x$, or you need to be allowed to take an infinite sum of functionals as you wrote above, where you have a non-zero contribution for every $n$. But also in this case, what you ask is morally true.