It's not a bad definition and I think it is better to think of it as a particular case of the "restriction problem", i.e., trying to define the restriction $\omega|_{\Gamma}$ of $\omega$ to some subset $\Gamma\subset\mathbb{R}^n$.
When one succeeds the result is called a *trace theorem*. This usually requires some quantitative regularity hypothesis on $\omega$, e.g., being in a Sobolev space $H^s$ with $s>$something.

A particularly important case is when $\Gamma$ is an affine subspace, or say for simplicity a linear subspace like $\Gamma =\mathbb{R}^m\times\{0\}^{n-m}\subset\mathbb{R}^n$.
A rather standard approach is to start with $\omega\in\mathcal{D}'(\mathbb{R}^n)$.
The convolution $\omega\ast \phi_j^0$ is in the space of $C^{\infty}$ functions
$\mathcal{E}(\mathbb{R}^n)\subset
\mathcal{D}'(\mathbb{R}^n)$ and converges to $\omega$
in the topology of $\mathcal{D}'(\mathbb{R}^n)$ (the strong topology).
The ordinary restriction $\omega\ast \phi_j^0|_{\Gamma}$ makes sense
and you can ask if the limit $\lim_{j\rightarrow\infty}\omega\ast \phi_j^0|_{\Gamma}$
exists inside $\mathcal{D}'(\mathbb{R}^m)$.

Your particular case $p=0$ corresponds to mine with $m=0$.

Another problem of this kind is pointwise multiplication. If $\omega_1(x)$ and $\omega_2(x)$ are two distributions, then there is no problem defining $\omega_1(x_1)\omega_2(x_2)$ (tensor product), but the issue is how to restrict to the diagonal $\Gamma=\{x_1=x_2\}$.

Finally, note that all of these problems become much more interesting for random distributions, because it's like magic: you can sometimes do the (deterministically) impossible.

**Small addendum:** Suppose that for some reason one has a trace theorem but only for large enough $m$ and one cannot do the $m=0$ or the point restriction case. Then one can still do the following "stabilization" trick: change $\omega$ to $\omega\otimes 1$ where one tensors with the constant function equal to one seen as a distribution in say $p$ new variables. If you can restrict it from $\mathbb{R}^{n+p}$ to a subspace of dimension $p$, then you will have your point evaluation after factoring out the $\otimes 1$.
The last step of course needs your restriction construction to be invariant/covariant by translation along $\Gamma$.

It's not a bad definition and I think it is better to think of it as a particular case of the "restriction problem", i.e., trying to define the restriction $\omega|_{\Gamma}$ of $\omega$ to some subset $\Gamma\subset\mathbb{R}^n$.
When one succeeds the result is called a *trace theorem*. This usually requires some quantitative regularity hypothesis on $\omega$, e.g., being in a Sobolev space $H^s$ with $s>$something.

A particularly important case is when $\Gamma$ is an affine subspace, or say for simplicity a linear subspace like $\Gamma =\mathbb{R}^m\times\{0\}^{n-m}\subset\mathbb{R}^n$.
A rather standard approach is to start with $\omega\in\mathcal{D}'(\mathbb{R}^n)$.
The convolution $\omega\ast \phi_j^0$ is in the space of $C^{\infty}$ functions
$\mathcal{E}(\mathbb{R}^n)\subset
\mathcal{D}'(\mathbb{R}^n)$ and converges to $\omega$
in the topology of $\mathcal{D}'(\mathbb{R}^n)$ (the strong topology).
The ordinary restriction $\omega\ast \phi_j^0|_{\Gamma}$ makes sense
and you can ask if the limit $\lim_{j\rightarrow\infty}\omega\ast \phi_j^0|_{\Gamma}$
exists inside $\mathcal{D}'(\mathbb{R}^m)$.

Your particular case $p=0$ corresponds to mine with $m=0$.

Another problem of this kind is pointwise multiplication. If $\omega_1(x)$ and $\omega_2(x)$ are two distributions, then there is no problem defining $\omega_1(x_1)\omega_2(x_2)$ (tensor product), but the issue is how to restrict to the diagonal $\Gamma=\{x_1=x_2\}$.

Finally, note that all of these problems become much more interesting for random distributions, because it's like magic: you can do the (deterministically) impossible.

It's not a bad definition and I think it is better to think of it as a particular case of the "restriction problem", i.e., trying to define the restriction $\omega|_{\Gamma}$ of $\omega$ to some subset $\Gamma\subset\mathbb{R}^n$.
When one succeeds the result is called a *trace theorem*. This usually requires some quantitative regularity hypothesis on $\omega$, e.g., being in a Sobolev space $H^s$ with $s>$something.

A particularly important case is when $\Gamma$ is an affine subspace, or say for simplicity a linear subspace like $\Gamma =\mathbb{R}^m\times\{0\}^{n-m}\subset\mathbb{R}^n$.
A rather standard approach is to start with $\omega\in\mathcal{D}'(\mathbb{R}^n)$.
The convolution $\omega\ast \phi_j^0$ is in the space of $C^{\infty}$ functions
$\mathcal{E}(\mathbb{R}^n)\subset
\mathcal{D}'(\mathbb{R}^n)$ and converges to $\omega$
in the topology of $\mathcal{D}'(\mathbb{R}^n)$ (the strong topology).
The ordinary restriction $\omega\ast \phi_j^0|_{\Gamma}$ makes sense
and you can ask if the limit $\lim_{j\rightarrow\infty}\omega\ast \phi_j^0|_{\Gamma}$
exists inside $\mathcal{D}'(\mathbb{R}^m)$.

Your particular case $p=0$ corresponds to mine with $m=0$.

Another problem of this kind is pointwise multiplication. If $\omega_1(x)$ and $\omega_2(x)$ are two distributions, then there is no problem defining $\omega_1(x_1)\omega_2(x_2)$ (tensor product), but the issue is how to restrict to the diagonal $\Gamma=\{x_1=x_2\}$.

Finally, note that all of these problems become much more interesting for random distributions, because it's like magic: you can sometimes do the (deterministically) impossible.

**Small addendum:** Suppose that for some reason one has a trace theorem but only for large enough $m$ and one cannot do the $m=0$ or the point restriction case. Then one can still do the following "stabilization" trick: change $\omega$ to $\omega\otimes 1$ where one tensors with the constant function equal to one seen as a distribution in say $p$ new variables. If you can restrict it from $\mathbb{R}^{n+p}$ to a subspace of dimension $p$, then you will have your point evaluation after factoring out the $\otimes 1$.
The last step of course needs your restriction construction to be invariant/covariant by translation along $\Gamma$.