In the terminology of Gel'fand, Generalized Functions Volume 1, a regular distribution, $\Lambda_f$, is one that is obtained from a locally integrable function, $f$, by integration of the function times a test function: $$\Lambda_f(\phi)=\int_{\mathbb{R}^n}\!f(x)\phi(x)\,dx\qquad(\phi\in C^\infty_c(\mathbb{R}^n)).$$ I know that the Dirac delta distribution, defined by $$\delta_p(\phi)=\phi(p)\qquad(\phi\in C^\infty_c(\mathbb{R}^n),\,p\in\mathbb{R}^n)$$ is not a regular distribution. For the $n=1$ case, there is a proof in Gel'fand on page 4 of Volume 1, there is this post: How to prove that the Dirac delta is not a function?, and it is not hard to prove using the $C^\infty$-Urysohn Lemma, even for general integer $n$. (Find a mollified sequence of $C^\infty_c(\mathbb{R}^n)$ functions converging pointwise to the characteristic function of $\{p\}$, use Dominated Convergence and the fact that the Lebesgue measure of $\{p\}$ is zero, to get a contradiction.)
I believe that it must be true as well for all derivatives of the delta distribution. Again, for the $n=1$ case, there is this website from Professor Tao: https://terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/ in which one finds:
Exercise 11 (Derivative of the delta function) Let
. Show that the functional
for all test functions
is a distribution which does not arise from either a locally integrable function or a Radon measure. (Note how it is important here that
I have been trying to prove that $\delta_p'$ is not regular in $\mathbb{R}^1$; that is, that there does not exist $f\in L^1_{\rm loc}(\mathbb{R})$ such that $\delta_p'=\Lambda_f$, hoping that such a proof would be generalizable to $\mathbb{R}^n$. However, I haven't had any luck even with $n=1$. So my simplified first question, setting $p=0$ and $n=1$, is "why can't $\delta'$ be $\Lambda_f$ for some $f\in L^1_{\rm loc}(\mathbb{R})$?" If that can be answered, then the next question is can it be extended to $\mathbb{R}^n$? Is it true that, given any multi-index $\alpha$, there is no $f\in L^1_{\rm loc}(\mathbb{R}^n)$ such that $D^\alpha\delta_p=\Lambda_f$?
In working on the $n=1$, $p=0$ case, I start with the obvious: \begin{equation}\tag{1} -\phi'(0)=-\delta(\phi')=\delta'(\phi)=\Lambda_f(\phi)=\int_{-\infty}^\infty\!f(x)\phi(x)\,dx \qquad(\phi\in C^\infty_c(\mathbb{R})), \end{equation} and look for a sequence of test functions that generate a contradiction. The trick of letting $\{\phi_k\}$ be a mollified sequence in $C^\infty_c(\mathbb{R})$ such that $\phi_k$ converges pointwise to $\chi_{\{0\}}$ doesn't seem to help because it appears that $\phi'_k(0)$ will likely have to approach $0$ and therefore not generate a contradiction. Letting $\{\phi_k\}$ just be a bump that narrows down to $0$ but whose height remains constant, also doesn't seem to work, since again, the derivatives will be $0$ at the origin. I also hoped I might be able to make use of the fact that if $\phi_1$ is a fixed element of $C^\infty_c(\mathbb{R})$ such that $$\int_{-\infty}^\infty\!\phi_1(x)\,dx=1,$$ then any $\phi\in C^\infty_c(\mathbb{R})$ can be expressed as \begin{equation}\tag{2} \phi=\phi_0+\Biggl[\int_{-\infty}^\infty\!\phi(x)\,dx\Biggr]\phi_1, \end{equation} where $\phi_0\in C^\infty_c(\mathbb{R})$ and is the derivative of a function in $C^\infty_c(\mathbb{R})$, as used by Gel'fand on page 40 of Volume 1; however, I couldn't find a way where that would help.
I want to thank @MaoWao and @copper.hat for their suggestions that lead to this answer. I will rephrase the question to be more general, and then give an answer based on their suggestions.
$\DeclareMathOperator{\supp}{supp}$ Show that, given the nonempty open set $\Omega\subseteq\mathbb{R}^n$, and given various constants, that the distribution \begin{equation}\tag{3} \Lambda=\sum_{i=1}^m\sum_{\lvert\alpha\rvert\leq N}c_{i,\alpha}D^\alpha\delta_{p_i}\in\mathscr{D}'(\Omega) \end{equation} is not regular. $\alpha$ is a multi-index, and the constants are integers $m>0$, $N\geq 0$, complex $\{c_{i,\alpha}\colon i=1,\dots,m,\,\lvert\alpha\rvert\leq N\}$ (for each $i$, not all $c_{i,\alpha}$ are zero), and $\{p_1,\dots,p_m\}\subseteq\Omega$.
Proof: Let $P=\{p_1,\dots,p_m\}$ and let $\Omega_P=\Omega-P$, an open subset of $\Omega$. Let a tilde over a function variable represent the zero-extension of that function from $\Omega_P$ to $\Omega$. To reach a contradiction, suppose that there is a function $f\in L^1_{\rm loc}(\Omega)$, such that $\Lambda=\Lambda_f$; that is, such that for every $\psi\in\mathscr{D}(\Omega)$, $$\Lambda(\psi)=\int_\Omega\!f\psi\,dm_\Omega\qquad(\psi\in\mathscr{D}(\Omega))$$ where $m$ is Lebesgue measure on $\mathbb{R}^n$ and $m_\Omega$ is the restriction of $m$ to (the Lebesgue measurable subsets of) $\Omega$. Let $f_P$ be the restriction of $f$ to $\Omega_P$. Then $f_P\in L^1_{\rm loc}(\Omega_P)$, for if $K$ is a compact subset of $\Omega_P$, then it is also a compact subset of $\Omega$ so that $\chi_Kf\in L^1(\Omega)$, and hence $\chi_Kf_P=(\chi_Kf)|_{\Omega_P}\in L^1(\Omega_P)$. Let $\phi\in\mathscr{D}(\Omega_P)$. Then $\tilde{\phi}\in\mathscr{D}(\Omega)$ since the support of $\phi$ is contained in $\Omega_P$. We have \begin{align*} \int_{\Omega_P}\!f_P\phi\,dm_{\Omega_P} &=\int_\Omega\!(f_P\phi)\tilde{\phantom{a}}\,dm_\Omega\\ &=\int_\Omega\!\chi_{\Omega_P}f\tilde{\phi}\,dm_\Omega\\ &=\int_\Omega\!f\tilde{\phi}\,dm_\Omega &\qquad&(\text{since $\supp\tilde{\phi}=\supp\phi\subseteq\Omega_P$})\\ &=\Lambda(\tilde{\phi})\\ &=\sum_{i=1}^m\sum_{\lvert\alpha\rvert\leq N}c_{i,\alpha}D^\alpha\delta_{p_i}(\tilde{\phi})\\ &=\sum_{i=1}^m\sum_{\lvert\alpha\rvert\leq N}(-1)^{\lvert\alpha\rvert}c_{i,\alpha}\delta_{p_i}(D^\alpha\tilde{\phi})\\ &=\sum_{i=1}^m\sum_{\lvert\alpha\rvert\leq N}(-1)^{\lvert\alpha\rvert}c_{i,\alpha}D^\alpha\tilde{\phi}(p_i)\\ &=0&&(\text{since $p_i\notin\supp\tilde{\phi}$}). \end{align*} Therefore, $f_P=0$ almost everywhere on $\Omega_P$. Since $P$ is a finite set, $f=0$ almost everywhere on $\Omega$, and therfore $\Lambda=0$. The inner sum in (3) is a nonzero distribution ($\Lambda_i$) whose support is $\{p_i\}$. If $\omega_i$ is an open neighborhood of $p_i$ which does not contain $p_j$ for $j\neq i$, then there is a $\phi\in\mathscr{D}(\Omega)$ whose support is contained in $\omega_i$ such that $\Lambda_i(\phi)\neq 0$. Furthermore, since the support of $\phi$ doesn't intersect the support of $\Lambda_j$ for $j\neq i$, we have that $\Lambda_j(\phi)=0$ for $j\neq i$, and so $\Lambda(\phi)=\Lambda_i(\phi)\neq 0$, a contradiction. $$\tag*{$\blacksquare$}$$