I’m going to reparameterize $a=\nu/2$.
If $p=1$ then we can take $\nu = 1 \implies a = \frac 12$ to get
$$
2^{\nu p/2}\Gamma_{p}(\nu/2) = \sqrt{2\pi} = (2\pi)^{p/2}
$$
so there does exist a $\nu$ such the two are equal in the trivial case of $p=1$.
In general, though, we have the following result:
Claim: for $p \in \{2, 3, \dots\}$ we have
$$
2^{ap} \Gamma_p(a) > (2\pi)^{p/2}
$$
for all $a>(p-1)/2$.
Proof: by induction on $p$.
Base case: $p=2$. I claim that
$$
f(a) := 2^{2a} \Gamma_2(a) - 2\pi > 0
$$
for $a>1/2$. We have the identity that
$$
\Gamma_2(a) = \pi^{1/2} \Gamma(a)\Gamma(a - 1/2)
$$
so $f(a) = 2^{2a}\pi^{1/2}\Gamma(a)\Gamma(a-1/2) - 2\pi$. Let $h(a) = 2^{2a}\pi^{1/2}\Gamma(a)\Gamma(a-1/2)$. Recalling that $\log \Gamma$ is convex, this means
$$
\log h(a) = a\log 4 + \frac 12 \log \pi + \log \Gamma(a) + \log \Gamma(a-1/2)
$$
is too, and $x\mapsto e^x$ is convex and non-decreasing, so we know $\exp \circ \log \circ h = h$ is convex. The function $x \mapsto -2\pi$ is also convex so this means that $f$ itself is convex.
I'm going to use the convexity of $f$ to construct a piecewise linear function $g$ such that $f \geq g > 0$. The linear pieces of $g$ will be supporting hyperplanes (ie tangent lines) of $f$. Letting $h_x(a)$ be the tangent line of $f$ at $x$, we have $h_x(a) = f(x) + f'(x)(a - x)$. By the convexity of $f$ I know that for all $x \in (1/2, \infty)$, $f \geq h_{x}$. Numerically you can find that the minimum of $f$ is at $a_0 \approx 1.23$, so I'm going to obtain $h_{1}$ and $h_{1.5}$ and then set $g(a) := \max\{h_1(a), h_{1.5}(a)\}$. It's important that $1 < a_0 < 1.5$ because this tells me that $h_1$ is decreasing while $h_{1.5}$ is increasing, so the minimum of $g$ is at their intersection. If I can show this is positive then I'm done.
In order to work out $h_1$ and $h_{1.5}$ we need $f'$, which using the product rule and substituting $\psi = \Gamma' / \Gamma$ we have
$$
f'(a) = 2^{2a}\Gamma_2(a)\left(\log 4 + \psi(a) + \psi(a-1/2)\right)
$$
where $\psi$ is the digamma function. It is well-known that $\Gamma(z+1) = z\Gamma(z)$ and $\psi(1 + z) = \frac 1z + \psi(z)$. We also have the special values $\Gamma(1/2) = \sqrt \pi$, $\psi(1) = -\gamma$, and $\psi(1/2) = -2\log 2 - \gamma$ where $\gamma$ is the Euler–Mascheroni constant. All together we can use those special values and the aforementioned identities to work out
$$
h_1(a) = 2\pi - 8\pi \gamma (a - 1)
$$
and
$$
h_{1.5}(a) = 2\pi + 8\pi(1-\gamma)(a - 1.5).
$$
These two lines intersect at $\alpha := \frac{3-\gamma}{2}$, and at that point we have
$$
g(\alpha) = 2\pi - 8\pi\gamma \left(\frac{3-\gamma}{2} - 1\right) > 0.
$$
This means that for all $a \in (1/2, \infty)$ we have $f(a) \geq g(a) > 0$ and that concludes the base case.
Inductive case: suppose true for $p=2,3,\dots, n$. Then in particular we have
$$
\Gamma_n(a) > \pi^{n/2}2^{n(1/2-a)}
$$
for all $a>(n-1)/2$ and we want to show
$$
\Gamma_{n+1}(a) \stackrel{\text{NTS}}> \pi^{(n+1)/2}2^{(n+1)(1/2-a)}
$$
for all $a>n/2$. Note that $1/2-a<0$ so $2^{1/2-a} < 1$. I'm going to use that $\stackrel{\text{NTS}}>$ (NTS = Need To Show) to denote inequalities that we are trying to prove but haven't yet.
From the linked Wikipedia article on the multivariate Gamma function we have
$$
\Gamma_{n+1}(a)=\pi^{n/2}\Gamma_n(a)\Gamma(a - n/2)
$$
$$
> \pi^n 2^{n(1/2-a)} \Gamma(a-n/2)
$$
$$
> \pi^n 2^{(n+1)(1/2-a)} \Gamma(a-n/2)
$$
so if we show
$$
\pi^n \Gamma(a-n/2) \stackrel{\text{NTS}}> \pi^{(n+1)/2}
$$
then we’re done.
This is equivalent to showing that
$$
\Gamma(a-n/2) \stackrel{\text{NTS}}> \pi^{(1-n)/2}.
$$
$a-n/2>0$ so the smallest that $\Gamma(a-n/2)$ can be is its unique minimum on the positive reals which is greater than $0.75$. But for $n\geq 3$ we have $\pi^{(1-n)/2}<1/3$ so no matter what for any valid $a$ we have $\Gamma(a-n/2) > \pi^{(1-n)/2}$.
Putting this all together, we have established the following chain of inequalities:
$$
\pi^{(n+1)/2}2^{(n+1)(1/2-a)} = \pi^{(n+1)/2}2^{n(1/2-a)}2^{1/2-a}
$$
$$
< \pi^{(n+1)/2}2^{n(1/2-a)} < \pi^n 2^{n(1/2-a)}\Gamma(a - n/2)
$$
$$
= \pi^{n/2} \left[\pi^{n/2} 2^{n(1/2-a)}\right]\Gamma(a - n/2) < \pi^{n/2} \Gamma_{n}(a) \Gamma(a - n/2)
$$
$$
= \Gamma_{n+1}(a) \implies 2^{a(n+1)}\Gamma_{n+1}(a) > (2\pi)^{(n+1)/2}
$$
for all $a > ((n+1)-1)/2$.
$\square$
